JEE Advanced Specific Heat Capacity & Thermodynamical Processes PYQ

Q 1 :

One mole of an ideal gas expands adiabatically from an initial state $(T_{A}, V_{0})$ to final state $(T_{f}, 5 V_{0})$ . Another mole of the same gas expands isothermally from a different initial state $(T_{B}, V_{0})$ to the same final state $(T_{f}, 5 V_{0})$ . The ratio of the specific heats at constant pressure and constant volume of this ideal gas is $γ$ . What is the ratio $T_{A} / T_{B}$ ? [2023]

$5^{γ - 1}$
$5^{1 - γ}$
$5^{γ}$
$5^{1 + γ}$

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(1)

$For adiabatic process$ $T V^{γ - 1} = constant$

$∴$ $T_{A} V_{0}^{γ - 1} = T_{f} {(5 V_{0})}^{γ - 1} ...(i)$

and for isothermal process, no change in temperature

$∴$ $T_{B} = T_{f} ...(ii)$

Dividing (i) by (ii)

[IMAGE 453]

$\frac{T_{A}}{T_{B}} = 5^{γ - 1}$

Q 2 :

An ideal gas is initially at $P_{1}, V_{1}$ is expanded to $P_{2}, V_{2}$ and then compressed adiabatically to the same volume $V_{1}$ and pressure $P_{3}$ . If $W$ is the net work done by the gas in complete process which of the following is true [2004]

$W > 0; P_{3} > P_{1}$
$W < 0; P_{3} > P_{1}$
$W > 0; P_{3} < P_{1}$
$W < 0; P_{3} < P_{1}$

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(2)

In the first process $W_{A B}$ is +ve as $Δ V$ is positive.

In the second process $W_{B C}$ is $-$ ve as $Δ V$ is $-$ ve and area under the curve of second process is more.

$∴$ $Net Work < 0$ and also $P_{3} > P_{1}$

[IMAGE 454]

Q 3 :

The PT diagram for an ideal gas is shown in the figure, where AC is an adiabatic process, find the corresponding PV diagram. [2003]

[IMAGE 455]

[IMAGE 456]
[IMAGE 457]
[IMAGE 458]
[IMAGE 459]

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(2)

From the P–T graph we find AB to be a isothermal process, AC is adiabatic process given. Also for an expansion process, the slope of adiabatic curve is more (or we can say that the area under the P–V graph for isothermal process is more than adiabatic process for same increase in volume). Hence graph (2) correctly depicts corresponding P–V graph.

Q 4 :

An ideal gas is taken through the cycle A → B → C → A, as shown in the figure. If the net heat supplied to the gas in the cycle is 5 J, the work done by the gas in the process C → A is [2002]

[IMAGE 460]

−5 J
−10 J
−15 J
−20 J

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(1)

$For cyclic process;$

$Q_{cyclic} = W_{A B} + W_{B C} + W_{C A} = 10 J + 0 + W_{C A} = 5 J$

$∴$ $W_{C A} = - 5 J$

Q 5 :

P-V plots for two gases during adiabatic processes are shown in the figure. Plots 1 and 2 should correspond respectively to [2001]

[IMAGE 461]

He and $O_{2}$
$O_{2}$ and He
He and Ar
$O_{2}$ and $N_{2}$

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(2)

$For adiabatic process P V^{γ} = constant$

Also for monoatomic gas like helium, $γ = 1.67$

for diatomic gas like oxygen, $γ = 1.4$

[IMAGE 462]

Since, $γ_{diatomic} < γ_{monoatomic}$

$∴$ $P_{diatomic} > P_{monoatomic}$

Hence graph 1 is for diatomic i.e., oxygen and graph 2 is for monoatomic i.e., for helium.

Q 6 :

A monoatomic ideal gas, initially at temperature $T_{1}$ , is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature $T_{2}$ by releasing the piston suddenly. If $L_{1}$ and $L_{2}$ are the length of the gas column before and after expansion respectively, then $\frac{T_{1}}{T_{2}}$ is given by [2000]

${(\frac{L_{1}}{L_{2}})}^{2 / 3}$
$\frac{L_{1}}{L_{2}}$
$\frac{L_{2}}{L_{1}}$
${(\frac{L_{2}}{L_{1}})}^{2 / 3}$

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(4)

$From T V^{γ - 1} = constant$

For monoatomic gas $γ = \frac{5}{3}$ , hence $T V^{2 / 3} = constant$

$∴$ $T_{1} L_{1}^{2 / 3} = T_{2} L_{2}^{2 / 3}$ $(∵ V \propto L)$

Hence, $\frac{T_{1}}{T_{2}} = {(\frac{L_{2}}{L_{1}})}^{2 / 3}$

Q 7 :

One mole of an ideal gas undergoes two different cyclic processes I and II, as shown in the P-V diagrams below. In cycle I, processes $a, b, c$ and $d$ are isobaric, isothermal, isobaric and isochoric, respectively. In cycle II, processes $a', b', c'$ and $d'$ are isothermal, isochoric, isobaric and isochoric, respectively. The total work done during cycle I is $W_{I}$ and that during cycle II is $W_{I I}$ . The ratio $\frac{W_{I}}{W_{I I}}$ is _______. [2023]

[IMAGE 463]

(2)

From P-V diagram

[IMAGE 464]

Work done during cycle I

$W_{I} = W_{a} + W_{b} + W_{c} + W_{d}$

$= 4 P_{0} (2 V_{0} - V_{0}) + n R T \ln (\frac{4 V_{0}}{2 V_{0}}) + 2 P_{0} (V_{0} - 4 V_{0}) + 0$

$= 4 P_{0} V_{0} + n R (\frac{8 P_{0} V_{0}}{n R}) \ln 2 - 6 P_{0} V_{0}$

$= 8 P_{0} V_{0} \ln 2 - 2 P_{0} V_{0}$

Work done during cycle II

$W_{I I} = W_{a}' + W_{b}' + W_{c}' + W_{d}'$

$= n R T \ln (\frac{2 V_{0}}{V_{0}}) + 0 + P_{0} (V_{0} - 2 V_{0}) + 0$

$= n R (\frac{4 P_{0} V_{0}}{n R}) \ln 2 - P_{0} V_{0}$ $= 4 P_{0} V_{0} \ln 2 - P_{0} V_{0}$

$∴$ $\frac{W_{I}}{W_{I I}} = \frac{8 P_{0} V_{0} \ln 2 - 2 P_{0} V_{0}}{4 P_{0} V_{0} \ln 2 - P_{0} V_{0}} = 2$

Q 8 :

An incompressible liquid is kept in a container having a weightless piston with a hole. A capillary tube of inner radius 0.1 mm is dipped vertically into the liquid through the airtight piston hole, as shown in the figure. The air in the container is isothermally compressed from its original volume $V_{0}$ to $\frac{100}{101} V_{0}$ with the movable piston. Considering air as an ideal gas, the height $(h)$ of the liquid column in the capillary above the liquid level in cm is __________.

[Given: Surface tension of the liquid is $0.075 {Nm}^{- 1}$ , atmospheric pressure is $10^{5} {Nm}^{- 2}$ , acceleration due to gravity $(g)$ is $10 {ms}^{- 2}$ , density of the liquid is $10^{3} {kg m}^{- 3}$ and contact angle of capillary surface with the liquid is zero] [2023]

[IMAGE 465]

(25)

[IMAGE 466]

$h_{0} = \frac{2 T \cos θ}{ρ g r} = \frac{2 T \cos 0 °}{ρ g r}$

$= \frac{2 \times 0.075 \times 1}{10^{3} \times 10 \times 10^{- 4}} = 15 cm$

$P_{0} V_{0} = P (\frac{100 V_{0}}{101}) \Rightarrow P = \frac{101}{100} P_{0}$

$P_{0} - \frac{2 T \cos θ}{r} + ρ g h = P = \frac{101}{100} P_{0}$

$\Rightarrow - ρ g h_{0} + ρ g h = \frac{P_{0}}{100} \Rightarrow h = h_{0} + \frac{P_{0}}{100 ρ g}$

$∴$ $h = 15 cm + \frac{10^{5} \times 100}{100 \times 10^{3} \times 10} = 25 cm$

Q 9 :

A diatomic ideal gas is compressed adiabatically to $\frac{1}{32}$ of its initial volume. If the initial temperature of the gas is $T_{i}$ (in Kelvin) and the final temperature is $a T_{i}$ , the value of $a$ is ______. [2010]

(4)

For an adiabatic process, applying $T V^{γ - 1} = constant$

$T_{1} V_{1}^{γ - 1} = T_{2} V_{2}^{γ - 1} \Rightarrow T_{1} = T_{2} {(\frac{V_{2}}{V_{1}})}^{γ - 1}$

$γ = 1.4 (for diatomic gas), V_{2} = \frac{V_{1}}{32}, T_{1} = T_{i}, T_{2} = a T_{i}$

$∴$ $T_{i} = a T_{i} {[\frac{1}{32}]}^{1.4 - 1} = a T_{i} {[\frac{1}{2^{5}}]}^{0.4} = \frac{a T_{i}}{4}$

$∴$ $a = 4$

Q 10 :

An ideal monoatomic gas of $n$ moles is taken through a cycle $W X Y Z W$ consisting of consecutive adiabatic and isobaric quasi-static processes, as shown in the schematic $V - T$ diagram. The volume of the gas at $W$ , $X$ and $Y$ points are $64 {cm}^{3}$ , $125 {cm}^{3}$ and $250 {cm}^{3}$ , respectively. If the absolute temperature of the gas $T_{W}$ at the point $W$ is such that $n R T_{W} = 1 J$ ( $R$ is the universal gas constant), then the amount of heat absorbed (in J) by the gas along the path $X Y$ is ______. [2025]

[IMAGE 467]

(1.6)

[IMAGE 468]

$Q_{x y} = P_{1} (V_{y} - V_{x}) + n R (T_{y} - T_{x}) \frac{3}{2}$

$Q_{x y} = \frac{5}{2} R (T_{y} - T_{x}) n$

$T_{w} \cdot {(64)}^{\frac{2}{3}} = T_{x} \cdot {(125)}^{\frac{2}{3}}$

$T_{w} \cdot 16 = T_{x} \cdot 25$

$T_{x} = \frac{16}{25} T_{w}$

$\frac{T_{y}}{V_{y}} = \frac{T_{x}}{V_{x}}$

$T_{y} = T_{x} (\frac{250}{125}) = 2 T_{x}$

$Q_{x y} = \frac{5}{2} n R (T_{x}) = \frac{5}{2} n R \cdot \frac{16}{25} T_{w}$ $[∵ n R T_{w} = 1 J given]$

$∴$ $Q_{x y} = \frac{8}{5} = 1.6$

Q 11 :

Consider one mole of helium gas enclosed in a container at initial pressure $P_{1}$ and volume $V_{1}$ . It expands isothermally to volume $4 V_{1}$ . After this, the gas expands adiabatically and its volume becomes $32 V_{1}$ . The work done by the gas during isothermal and adiabatic expansion processes are $W_{iso}$ and $W_{adia}$ , respectively. If the ratio $\frac{W_{iso}}{W_{adia}} = f \ln 2,$ then $f$ is _______. [2020]

(1.78)

[IMAGE 469]

$For monoatomic gas, γ = \frac{5}{3}$

$In adiabatic process$ $P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ}$

$\Rightarrow \frac{P_{1}}{4} {(4 V_{1})}^{5 / 3} = P_{2} {(32 V_{1})}^{5 / 3}$

$\Rightarrow P_{2} = \frac{P_{1}}{4} {(\frac{1}{8})}^{5 / 3} = \frac{P_{1}}{128}$

$And W_{adi} = \frac{P_{1} V_{1} - P_{2} V_{2}}{γ - 1} = \frac{P_{1} V_{1} - \frac{P_{1}}{128} (32 V_{1})}{\frac{5}{3} - 1}$

$= \frac{P_{1} V_{1} (\frac{3}{4})}{\frac{2}{3}} = \frac{9}{8} P_{1} V_{1}$

$In isothermal process,$

$W_{iso} = 2.303 μ R T \log_{10} (\frac{V_{2}}{V_{1}})$

$\Rightarrow W_{iso} = P_{1} V_{1} \ln (\frac{4 V_{1}}{V_{1}}) = 2 P_{1} V_{1} \ln 2$

$∴$ $\frac{W_{iso}}{W_{adia}} = \frac{2 P_{1} V_{1} \ln 2}{\frac{9}{8} P_{1} V_{1}} = \frac{16}{9} \ln 2 = f \ln 2$

$So, f = \frac{16}{9} = 1.7778 \approx 1.78$

Q 12 :

A spherical bubble inside water has radius $R$ . Take the pressure inside the bubble and the water pressure to be $p_{0}$ . The bubble now gets compressed radially in an adiabatic manner so that its radius becomes $(R - a)$ . For $a ≪ R$ , the magnitude of the work done in the process is given by $(4 π p_{0} R a^{2}) X$ where $X$ is a constant and $γ = \frac{C_{p}}{C_{v}} = \frac{41}{30} .$ The value of $X$ is _______. [2020]

(2.05)

$Here, we will neglect surface tension.$

$As P_{inside} = P_{outside} = P_{0}$

$Here, Δ k = 0$

$So, W_{all forces} = 0$

$W_{ext} + W_{w} + W_{g} = 0, W_{w} \to work done by water$

$W_{ext} = - (W_{w} + W_{g}), W_{g} \to work done by gas$

$Now, W_{w} = P_{0} [\frac{4}{3} π R^{3} - \frac{4}{3} π {(R - a)}^{3}]$

$= \frac{4}{3} P_{0} π [R - (R - a)] [R^{2} + R (R - a) + {(R - a)}^{2}]$

$= \frac{4}{3} P_{0} π (a) [R^{2} + R^{2} + R^{2} - 3 R a + a^{2}]$

$= 4 P_{0} π [R^{2} a - R a^{2}]$

$and,$ $W_{g} = \frac{P_{1} V_{1} - P_{2} V_{2}}{γ - 1} = P_{0} \frac{4}{3} π R^{3} - P_{0} {(\frac{R}{R - a})}^{41 / 10} \frac{4}{3} π {(R - a)}^{3}$ $[∵ P V^{γ} = constant]$

$= \frac{30}{11} P_{0} \times \frac{4}{3} π R^{3} [1 - {(\frac{R}{R - a})}^{41 / 10} {(\frac{R - a}{R})}^{3}]$

$= \frac{40}{11} P_{0} π R^{3} [1 - {(1 - \frac{a}{R})}^{- 11 / 10}]$

$= \frac{40}{11} P_{0} π R^{3} [1 - 1 - \frac{11 a}{10 R} - \frac{(- \frac{11}{10}) (- \frac{21}{10})}{2} \frac{a^{2}}{R^{2}}]$

$= - 4 P_{0} π R^{2} a - 4.2 P_{0} π R a^{2}$

$So, W_{w} + W_{g} = - 4 P_{0} π R_{0}^{2} [1 + 1.05] = - 4 P_{0} π R_{0}^{2} \times 2.05$

$∴$ $W_{ext} = 4 P_{0} π R a^{2} \times 2.05$

$So, X = 5$

Q 13 :

An ideal gas of density $ρ = 0.2 {kg m}^{- 3}$ enters a chimney of height $h$ at the rate of $α = 0.8 {kg s}^{- 1}$ from its lower end, and escapes through the upper end as shown in the figure. The cross-sectional area of the lower end is $A_{1} = 0.1 m^{2}$ and the upper end is $A_{2} = 0.4 m^{2} .$ The pressure and the temperature of the gas at the lower end are $600 Pa$ and $300 K$ , respectively, while its temperature at the upper end is $150 K$ . The chimney is heat insulated so that the gas undergoes adiabatic expansion. Take $g = 10 {m s}^{- 2}$ and the ratio of specific heats of the gas $γ = 2 .$ Ignore atmospheric pressure.

[IMAGE 470]

Which of the following statement(s) is(are) correct? [2022]

The pressure of the gas at the upper end of the chimney is $300 Pa$ .
The velocity of the gas at the lower end of the chimney is $40 {m s}^{- 1}$ and at the upper end is $20 {m s}^{- 1}$ .
The height of the chimney is $590 m$ .
The density of the gas at the upper end is $0.05 {kg m}^{- 3}$ .

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(2)

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$As expansion of gas is adiabatic$

$so, P T^{\frac{γ}{1 - γ}} = constant$

$\Rightarrow 600 \times 300^{- 2} = P_{2} \times 150^{- 2}$

$\Rightarrow P_{2} = \frac{600 \times 150^{2}}{300^{2}}$

$\Rightarrow P_{2} = 150 Pa$

$So (1) is incorrect.$

$As \frac{d m}{d t} = ρ A v$

$At 1.$

$0.8 = 0.2 \times 0.1 \times v_{1}$

$v_{1} = 40 m/s$

$By ideal gas equation$

$P V = n R T$

$\Rightarrow P V = \frac{m}{M_{0}} R T \Rightarrow P M_{0} = ρ R T \Rightarrow \frac{ρ}{P T} = constant$

$\Rightarrow \frac{P_{1}}{ρ_{1} T_{1}} = \frac{P_{2}}{ρ_{2} T_{2}} \Rightarrow \frac{600}{0.2 \times 300} = \frac{150}{ρ_{2} \times 150}$

$\Rightarrow ρ = 0.1 {kg/m}^{3}$

$So option (4) is incorrect.$

$Again, \frac{d m}{d t} = ρ A v$

$At 2.$

$0.8 = 0.1 \times 0.4 \times v_{2}$

$\Rightarrow v_{2} = 20 m/s$

$So option (2) is correct.$

$Here, T \neq constant. So Bernoulli's theorem needs to be modified.$

$So, we need to add internal energy per unit volume factor.$

$Therefore, factor of \frac{n C_{v} T}{V} is added.$

$As \frac{n C_{v} T}{V} = \frac{n (\frac{f}{2} R) T}{V} = \frac{n R T}{V} \times \frac{2}{2} = P$

$So, Bernoulli's theorem becomes$ $2 P + \frac{1}{2} ρ v^{2} + ρ g h = constant$

$\Rightarrow 2 P_{1} + \frac{1}{2} ρ_{1} v_{1}^{2} + ρ_{1} g \times 0 = 2 P_{2} + \frac{1}{2} ρ_{2} v_{2}^{2} + ρ_{2} g h$

$\Rightarrow 2 \times 600 + \frac{1}{2} \times 0.2 \times 40^{2} = 2 \times 150 + \frac{1}{2} \times 0.1 \times 20^{2} + 0.1 \times 10 \times h$

$\Rightarrow h = 1040 m$

$So option (3) is incorrect.$

Q 14 :

A bubble has surface tension $S$ . The ideal gas inside the bubble has ratio of specific heats $γ = \frac{5}{3} .$ The bubble is exposed to the atmosphere and it always retains its spherical shape. When the atmospheric pressure is $P_{a 1}$ , the radius of the bubble is found to be $r_{1}$ and the temperature of the enclosed gas is $T_{1}$ . When the atmospheric pressure is $P_{a 2}$ , the radius of the bubble and the temperature of the enclosed gas are $r_{2}$ and $T_{2}$ , respectively.

Which of the following statement(s) is(are) correct? [2022]

If the surface of the bubble is a perfect heat insulator, then ${(\frac{r_{1}}{r_{2}})}^{5} = \frac{P_{a 2} + \frac{2 S}{r_{2}}}{P_{a 1} + \frac{2 S}{r_{1}}} .$
If the surface of the bubble is a perfect heat insulator, then the total internal energy of the bubble including its surface energy does not change with the external atmospheric pressure.
If the surface of the bubble is a perfect heat conductor and the change in atmospheric temperature is negligible, then ${(\frac{r_{1}}{r_{2}})}^{3} = \frac{P_{a 2} + \frac{4 S}{r_{2}}}{P_{a 1} + \frac{4 S}{r_{1}}} .$
If the surface of the bubble is a perfect heat insulator, then ${(\frac{T_{2}}{T_{1}})}^{\frac{5}{2}} = \frac{P_{a 2} + \frac{4 S}{r_{2}}}{P_{a 1} + \frac{4 S}{r_{1}}} .$

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Select one or more options

(3, 4)

$If the surface of the bubble is perfect heat insulator,$

$So process is adiabatic, so,$

$P V^{γ} = constant$

$\Rightarrow (P_{a 1} + \frac{4 s}{r_{1}}) {(\frac{4}{3} π r_{1}^{3})}^{5 / 3} = (P_{a 2} + \frac{4 s}{r_{2}}) {(\frac{4}{3} π r_{2}^{3})}^{5 / 3}$

$\Rightarrow {(\frac{r_{1}}{r_{2}})}^{5} = \frac{P_{a 2} + \frac{4 s}{r_{2}}}{P_{a 1} + \frac{4 s}{r_{1}}}$

$So (1) is incorrect.$

$Also, T_{1} V_{1}^{γ - 1} = T_{2} V_{2}^{γ - 1}$

$\Rightarrow \frac{T_{2}}{T_{1}} = {(\frac{V_{1}}{V_{2}})}^{2 / 3} \Rightarrow \frac{T_{2}}{T_{1}} = {(\frac{r_{1}}{r_{2}})}^{2} \Rightarrow$ $\frac{T_{2}}{T_{1}}$ $= {(\frac{P_{a 2} + \frac{4 s}{r_{2}}}{P_{a 1} + \frac{4 s}{r_{1}}})}^{2 / 5}$

$So (4) is correct.$

$Total internal energy + surface energy will not be same as$ $work done by gas will be there$

$i.e. Δ U = - Δ W$ $(∵ Δ Q = 0)$

$So, (2) is incorrect.$

$Now, if bubble is perfect heat conductor, then temperature$ $will remain constant. So, P V = constant$

$(P_{a 1} + \frac{4 s}{r_{1}}) \frac{4}{3} π r_{1}^{3} = (P_{a 2} + \frac{4 s}{r_{2}}) \frac{4}{3} π r_{2}^{3}$

${(\frac{r_{1}}{r_{2}})}^{3} = (\frac{P_{a 2} + \frac{4 s}{r_{2}}}{P_{a 1} + \frac{4 s}{r_{1}}})$

$So (3) is correct.$

Q 15 :

In the given $P - V$ diagram, a monoatomic gas $(γ = \frac{5}{3})$ is first compressed adiabatically from state $A$ to state $B$ . Then it expands isothermally from state $B$ to state $C$ .

$[Given {(\frac{1}{3})}^{0.6} = 0.5, \ln 2 = 0.7]$

[IMAGE 472]

Which of the following statement(s) is(are) correct? [2022]

The magnitude of the total work done in the process $A \to B \to C$ is $144 k J$ .
The magnitude of the work done in the process $B \to C$ is $84 k J$ .
The magnitude of the work done in the process $A \to B$ is $60 k J$ .
The magnitude of the work done in the process $C \to A$ is zero.

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Select one or more options

(2, 3, 4)

$(1) W_{A \to B \to C} = W_{A \to B} + W_{B \to C}$

$= \frac{P_{A} V_{A} - P_{B} V_{B}}{γ - 1} + P_{A} V_{A} \ln (\frac{V_{C}}{V_{B}})$

$As, P_{A} V_{A}^{γ} = P_{B} V_{B}^{γ} \Rightarrow {(\frac{V_{B}}{V_{A}})}^{γ} = \frac{P_{A}}{P_{B}}$

$\Rightarrow V_{B} = {(\frac{P_{A}}{P_{B}})}^{1 / γ} V_{A} = 0.8 {(\frac{1}{3})}^{3 / 5} = 0.8 \times 0.5 = 0.4 m^{3}$

$So, W_{A \to B \to C} = \frac{100 \times 0.8 - 300 \times 0.4}{\frac{5}{3} - 1} + 300 \times 0.4 \ln (\frac{0.8}{0.4})$

$= \frac{- 40}{2 / 3} + 120 \ln 2 = - 60 + 120 \times 0.7 = - 60 + 84 = 24 kJ$

$So, (2,3,4) are correct.$

Q 16 :

A mixture of ideal gas containing 5 moles of monoatomic gas and 1 mole of rigid diatomic gas is initially at pressure $P_{0}$ , volume $V_{0}$ , and temperature $T_{0}$ . If the gas mixture is adiabatically compressed to a volume $\frac{V_{0}}{4},$ then the correct statement(s) is/are,

(Given $2^{1.2} = 2.3, 2^{3.2} = 9.2,$ R is gas constant) [2019]

The work $| W |$ done during the process is $13 R T_{0}$ .
The final pressure of the gas mixture after compression is in between $9 P_{0}$ and $10 P_{0}$ .
The average kinetic energy of the gas mixture after compression is in between $18 R T_{0}$ and $19 R T_{0}$ .
Adiabatic constant of the gas mixture is 1.6.

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Select one or more options

(1, 2, 4)

$Adiabatic constant of the gas mixture,$

$γ_{m} = \frac{n_{1} C_{p 1} + n_{2} C_{p 2}}{n_{1} C_{v 1} + n_{2} C_{v 2}} = \frac{5 \times \frac{5 R}{2} + 1 \times \frac{7 R}{2}}{5 \times \frac{3 R}{2} + 1 \times \frac{5 R}{2}} = 1.6$

$For an adiabatic process, P V^{γ} = Constant$

$∴$ $P = P_{0} {(\frac{V_{0}}{V})}^{1.6} = P_{0} {(4)}^{1.6}$

$= P_{0} {(2^{2})}^{1.6} = P_{0} 2^{3.2} = 9.2 P_{0}$

$Work done during the process,$

$W = \frac{P_{2} V_{2} - P_{1} V_{1}}{1 - γ} = \frac{9.2 P_{0} \times (\frac{V_{0}}{4}) - P_{0} V_{0}}{1 - 1.6} = \frac{- 13 P_{0} V_{0}}{6}$

$But P_{0} V_{0} = 6 R T_{0}$ $(as n = 5 + 1 = 6)$

$∴$ $W = \frac{- 13 (6 R T_{0})}{6} = - 13 R T_{0}$

$∴$ $| W |$ $= 13 R T_{0}$

$The average K.E. of the gas mixture,$

$\bar{K.E.} = n C V_{mi} \times T_{2}$

$From, T_{1} V_{1}^{γ - 1} = T_{2} V_{2}^{γ - 1}$

$or, T_{2} = T_{1} {(2)}^{6 / 5} = 23 T_{0}$

$∴$ $\bar{K.E.} = n C V_{mi} \times T_{2} = 23 R T_{0}$

Q 17 :

One mole of a monatomic ideal gas goes through a thermodynamic cycle, as shown in the volume versus temperature $(V - T)$ diagram. The correct statement(s) is/are: [R is the gas constant] [2019]

[IMAGE 473]

Work done in this thermodynamic cycle $(1 \to 2 \to 3 \to 4 \to 1)$ is $| W |$ $= \frac{1}{2} R T_{0}$
The above thermodynamic cycle exhibits only isochoric and adiabatic processes.
The ratio of heat transfer during processes $1 \to 2$ and $2 \to 3$ is $| \frac{Q_{1 \to 2}}{Q_{2 \to 3}} |$ $= \frac{5}{3}$
The ratio of heat transfer during processes $1 \to 2$ and $3 \to 4$ is $| \frac{Q_{1 \to 2}}{Q_{3 \to 4}} |$ $= \frac{1}{2}$

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Select one or more options

(1, 3)

$The P - V graph of the given V - T graph is given below.$

[IMAGE 474]

$(1) Work done during cyclic process (1 \to 2 \to 3 \to 4 \to 1)$

$W = area enclosed in the loop = \frac{P_{0}}{2} V_{0}$

$∵$ $P_{0} V_{0} = n R T_{0}$ $∴$ $\frac{P_{0} V_{0}}{2} = \frac{n R T_{0}}{2}$

$∴$ $Work done W = \frac{n R T_{0}}{1} = \frac{R T_{0}}{2}$ $[as n = 1]$

$(2) Process 1 \to 2 is isobaric$

$Process 2 \to 3 is isochoric$

$Process 3 \to 4 is isobaric$

$Process 4 \to 1 is isochoric$

$Hence no adiabatic process is involved.$

$(3) | Δ Q_{1 \to 2} | = | n C_{p} Δ T | = | n C_{p} (2 T_{0} - T_{0}) | = | n C_{p} T_{0} |$

$| Δ Q_{2 \to 3} | = | Δ U | = | n C_{v} Δ T | = | n C_{v} T_{0} |$

$∴$ $| \frac{Δ Q_{1 \to 2}}{Δ Q_{2 \to 3}} |$ $= \frac{C_{p}}{C_{v}} = \frac{5}{3}$

$(4) | Δ Q_{3 \to 4} | = n C_{p} \frac{T_{0}}{2}$

$∴$ $| \frac{Δ Q_{1 \to 2}}{Δ Q_{3 \to 4}} |$ $= \frac{n C_{p} T_{0}}{n C_{p} \frac{T_{0}}{2}} = \frac{2}{1}$

Q 18 :

One mole of an ideal gas in initial state $A$ undergoes a cyclic process $A B C A$ , as shown in the figure. Its pressure at $A$ is $P_{0}$ . Choose the correct option(s) from the following. [2010]

[IMAGE 475]

Internal energies at A and B are the same.
Work done by the gas in process AB is $P_{0} V_{0} \ln 4$
Pressure at C is $\frac{P_{0}}{4}$
Temperature at C is $\frac{T_{0}}{4}$

1

2

3

4

Select one or more options

(1, 2)

[IMAGE 476]

$Process A to B$

$Here, T_{A} = T_{B}$ $∴$ $U_{A} = U_{B}$

$Also P_{0} V_{0} = P_{B} \times 4 V_{0}$

$\Rightarrow P_{B} = \frac{P_{0}}{4}$

$Work done$

$W_{A B} = n R T_{0} \log_{e} (\frac{4 V_{0}}{V_{0}})$

$= P_{0} V_{0} \log_{e} 4$ $[∵ P_{0} V_{0} = n R T_{0}]$

$The process BC is not clear. Therefore no judgement can$ $be made at C regarding P and T .$

Q 19 :

The figure shows the $P - V$ plot of an ideal gas taken through a cycle $A B C D A$ . The part $A B C$ is a semi-circle and $C D A$ is half of an ellipse. Then, [2009]

[IMAGE 477]

The process during the path $A \to B$ is isothermal.
Heat flows out of the gas during the path $B \to C \to D$ .
Work done during the path $A \to B \to C$ is zero.
Positive work is done by the gas in the cycle $A B C D A$ .

1

2

3

4

Select one or more options

(2, 4)

$(1) Process A \to B is not isothermal. In case of an isothermal$ $process we get a rectangular hyperbola in a P - V diagram.$

$(2) In process B \to C \to D, Δ U is negative. P V decreases and$ $volume also decreases, therefore W is negative.$

$From first law$ $of thermodynamic, Q is negative i.e., there is a heat loss.$

$(3) W_{A B} > W_{B C} .$ $Therefore work done during path A \to B \to C is positive.$

$(4) Work done in clockwise cycle in a P - V diagram is positive.$

Q 20 :

One mole of a monatomic ideal gas undergoes the cyclic process $J \to K \to L \to M \to J$ , as shown in the $P - T$ diagram.

[IMAGE 478]

Match the quantities mentioned in List-I with their values in List-II and choose the correct option. [R is the gas constant] [2024]

List - I List - II

(P) Work done in the complete cyclic process (1) $R T_{0} - 4 R T_{0} \ln 2$

(Q) Change in the internal energy of the gas in the process JK (2) $0$

(R) Heat given to the gas in the process KL (3) $3 R T_{0}$

(S) Change in the internal energy of the gas in the process MJ (4) $- 2 R T_{0} \ln 2$

(5) $- 3 R T_{0} \ln 2$

P → 1; Q → 3; R → 5; S → 4
P → 4; Q → 3; R → 5; S → 2
P → 4; Q → 1; R → 2; S → 2
P → 2; Q → 5; R → 3; S → 4

1

2

3

4

(2)

$From given P - T diagram,$

$For process J K : Isobaric$ $W = n R (3 T_{0} - T_{0}) = 2 R T_{0}$

$and Δ u = n c_{v} Δ T = n \times \frac{3 R}{2} \times 2 T_{0} = 3 R T_{0}$

$For process K L : Isothermal$ $W = - n R (3 T_{0}) \ln 2$

$W = - 3 R T_{0} \ln 2$

$Δ u = 0 and Q = - 3 R T_{0} \ln 2$

$For process L M : Isobaric$

$W = n R (T_{0} - 3 T_{0}) = - 2 R T_{0}$

$Δ u = n c_{v} Δ T = n \times \frac{3 R}{2} \times (- 2 T_{0}) = - 3 R T_{0}$

$For process M J : Isothermal$

$W = n R T \ln 2 = R T_{0} \ln 2$

$Δ u = 0$

$And work done in the complete cyclic process,$

$W_{net} = R T_{0} \ln 2 + 2 R T_{0} - 3 R T_{0} \ln 2 - 2 R T_{0}$

$= - 2 R T_{0} \ln 2$

$So correct option is (2)$

Q 21 :

List I describes thermodynamic processes in four different systems. List II gives the magnitudes (either exactly or as a close approximation) of possible changes in the internal energy of the system due to the process. [2022]

List - I List - II

(I) $10^{- 3} kg$ of water at $100 ° C$ is converted to steam at the same temperature, at a pressure of $10^{5} Pa$ . The volume of the system changes from $10^{- 6} m^{3}$ to $10^{- 3} m^{3}$ in the process. Latent heat of water $= 2250 kJ/kg$ . (P) $2 kJ$

(II) $0.2$ moles of a rigid diatomic ideal gas with volume $V$ at temperature $500 K$ undergoes an isobaric expansion to volume $3 V$ . Assume $R = 8.0 {J mol}^{- 1} K^{- 1}$ . (Q) $7 kJ$

(III) One mole of a monatomic ideal gas is compressed adiabatically from volume $V = \frac{1}{3} m^{3}$ and pressure $2 kPa$ to volume $\frac{V}{8}$ . (R) $4 kJ$

(IV) Three moles of a diatomic ideal gas whose molecules can vibrate, is given $9 kJ$ of heat and undergoes isobaric expansion. (S) $5 kJ$

(T) $3 kJ$

Which one of the following options is correct?

(I) → (T); (II) → (R); (III) → (S); (IV) → (Q)
(I) → (S); (II) → (P); (III) → (T); (IV) → (P)
(I) → (P); (II) → (R); (III) → (T); (IV) → (Q)
(I) → (Q); (II) → (R); (III) → (S); (IV) → (T)

1

2

3

4

(3)

$(i) By first law of thermodynamics,$

$Δ U = Δ Q - Δ W = M L_{v} - P Δ V$

$= 10^{- 3} \times 2250 \times 10^{3} - 10^{5} \times (10^{- 3} - 10^{- 6})$

$= 2250 - 100 ≃ 2150 J = 2.15 kJ$ $So (I) \to (P)$

$(i i) P = \frac{n R T}{V} = \frac{0.2 \times 8 \times 500}{V} = \frac{800}{V} Pa$

$Δ U = \frac{f}{2} P Δ V = \frac{5}{2} \times \frac{800}{V} \times 2 V = 4000 J = 4 kJ$

$So (II) \to (R)$

$(i i i) P V^{γ} = constant \Rightarrow P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ}$

$\Rightarrow 2 V^{γ} = P_{2} {(\frac{V}{8})}^{γ} \Rightarrow P_{2} = 2 \times 8^{γ} = 2 \times 8^{5 / 3} = 64 kPa$

$So, Δ U = \frac{f}{2} (P_{2} V_{2} - P_{1} V_{1})$ $= \frac{3}{2} (64 \times \frac{1}{24} - 2 \times \frac{1}{3}) \times 10^{3}$ $= 3 kJ$

$So (III) \to (T)$

$(i v) Here f = 7$

$So, Δ U = n C_{v} Δ T = \frac{f}{2} n R Δ T = \frac{7}{2} n R Δ T$

$and, Δ Q = n C_{p} Δ T = (\frac{f}{2} + 1) n R Δ T = \frac{9}{2} n R Δ T = \frac{9}{2} \times \frac{2}{7} Δ U$

$Δ U = \frac{7}{9} Δ Q$

$So, Δ U = \frac{7}{9} Δ Q = \frac{7}{9} \times 9 = 7 kJ$

$So (IV) \to (Q)$

Q 22 :

If the process carried out on one mole of monatomic ideal gas is shown in the figure $P V$ -diagram with $P_{0} V_{0} = \frac{1}{3} R T_{0},$ the correct match is, [2019]

[IMAGE 479]

I → Q, II → R, III → S, IV → U
I → S, II → R, III → Q, IV → T
I → Q, II → R, III → P, IV → U
I → Q, II → S, III → R, IV → U

1

2

3

4

(1)

$I. W_{1 \to 2 \to 3} = W_{1 \to 2} + W_{2 \to 3}$ $[W_{2 \to 3} = 0, Δ V = 0]$

$= P_{0} \times V_{0} + 0$

$= P_{0} V_{0} = \frac{R T_{0}}{3}$ $[∵ P_{0} V_{0} = \frac{1}{3} R T_{0} given]$

$∴$ $I \to Q$

$II. Δ U_{1 \to 2 \to 3} = Δ U_{1 \to 2} + Δ U_{2 \to 3}$

$= n C_{v} Δ T_{1 \to 2} + n C_{v} Δ T_{2 \to 3}$

$= 1 \times \frac{3 R}{2} {(T_{f} - T_{i})}_{1 \to 2} + 1 \times \frac{3 R}{2} {(T_{f} - T_{i})}_{2 \to 3}$

$= \frac{3}{2} [2 P_{0} V_{0} - P_{0} V_{0}] + \frac{3}{2} [\frac{3 P_{0}}{2} \times 2 V_{0} - P_{0} \times 2 V_{0}]$

$= 3 P_{0} V_{0} = 3 \times \frac{1}{3} R T_{0} = R T_{0}$

$II \to R$

$III. Q_{1 \to 2 \to 3} = Q_{1 \to 2} + Q_{2 \to 3} = n C_{p} Δ T_{1 \to 2} + n C_{v} Δ T_{2 \to 3}$

$= \frac{5}{2} P_{0} V_{0} + 1 \times \frac{3}{2} [\frac{3 P_{0}}{2} \times 2 V_{0} - P_{0} (2 V_{0})]$

$= \frac{8}{2} P_{0} V_{0} = \frac{8}{2} \times \frac{R T_{0}}{3} = \frac{4}{3} R T_{0}$

$III \to S$

$IV. Q_{1 \to 2} = n C_{p} Δ T_{1 \to 2} = n C_{p} (T_{f} - T_{i})$

$= 1 \times \frac{5 R}{2} [\frac{P_{0} (2 V_{0})}{R} - \frac{P_{0} V_{0}}{R}]$ $[∵ P V = n R T]$

$= \frac{5}{2} P_{0} V_{0} = \frac{5}{2} (\frac{R T_{0}}{3}) = \frac{5 R T_{0}}{6}$

$∴$ $IV \to U$

Q 23 :

If the process carried out on one mole of monatomic ideal gas is as shown in the TV-diagram with $P_{0} V_{0} = \frac{1}{3} R T_{0},$ the correct match is, [2019]

[IMAGE 480]

I → P, II → T, III → Q, IV → T
I → S, II → T, III → Q, IV → U
I → P, II → R, III → T, IV → P
I → P, II → R, III → T, IV → S

1

2

3

4

(3)

$I. W_{1 \to 2 \to 3} = W_{1 \to 2} + W_{2 \to 3}$

$= n R T \ln (\frac{V_{f}}{V_{i}}) + P d V$

$= 1 R \frac{T_{0}}{3} \ln (\frac{2 V_{0}}{V_{0}}) + zero$ $[∵ n = 1, and W_{2 \to 3} isochoric no change in volume \Rightarrow Δ V = 0]$

$= \frac{R T_{0}}{3} \ln 2$ $∴$ $I \to P$

$II. Δ U_{1 \to 2 \to 3} = Δ U_{1 \to 2} + Δ U_{2 \to 3}$

$= 0 + n C_{v} Δ T = n \frac{f}{2} R Δ T$ $[Process 1 \to 2 isothermal, Δ T = 0]$

$= 1 \times \frac{3 R}{2} (T_{0} - \frac{T_{0}}{3}) = R T_{0}$ $[For monoatomic gas f = 3]$

$∴$ $II \to R$

$III. Q_{1 \to 2 \to 3} = W_{1 \to 2 \to 3} + Δ U_{1 \to 2 \to 3}$

$From first law of thermodynamics$

$= \frac{R T_{0}}{3} \ln 2 + R T_{0} = \frac{R T_{0}}{3} [\ln 2 + 3]$

$∴$ $III \to T$

$IV. Q_{1 \to 2} = W_{1 \to 2} + Δ U_{1 \to 2}$

$= \frac{R T_{0}}{3} \ln 2 + 0 = \frac{R T_{0}}{3} \ln 2$ $[∵ Δ U_{1 \to 2} = 0]$

$∴$ $IV \to P$

Q 24 :

One mole of a monatomic ideal gas undergoes four thermodynamic processes as shown schematically in the PV-diagram below. Among these four processes, one is isobaric, one is isochoric, one is isothermal and one is adiabatic. Match the processes mentioned in List-I with the corresponding statements in List-II. [2018]

[IMAGE 481]

LIST-I LIST-II

P. In process I 1. Work done by the gas is zero

Q. In process II 2. Temperature of the gas remains unchanged

R. In process III 3. No heat is exchanged between the gas and its surroundings

S. In process IV 4. Work done by the gas is $6 P_{0} V_{0}$

P → 4; Q → 3; R → 1; S → 2
P → 1; Q → 3; R → 2; S → 4
P → 3; Q → 4; R → 1; S → 2
P → 3; Q → 4; R → 2; S → 1

1

2

3

4

(3)

$Process I is adiabatic therefore Δ Q = 0$

$Process II is isobaric P = constant therefore$ $W = P (V_{2} - V_{1}) = 3 P_{0} (3 V_{0} - V_{0}) = 6 P_{0} V_{0}$

$Process III is isochoric V = constant therefore$ $W = P (V_{2} - V_{1}) = 0$

$Process IV is isothermal, temperature T = constant$

$∴$ $Δ u = 0$

Q 25 :

By appropriately matching the information given in the three columns of the following table. An ideal gas is undergoing a cyclic thermodynamic process in different ways as shown in the corresponding P–V diagrams in column 3 of the table. Consider only the path from state 1 to state 2. W denotes the corresponding work done on the system. The equations and plots in the table have standard notations as used in thermodynamic processes. Here Y is the ratio of heat capacities at constant pressure and constant volume. The number of moles in the gas is $n$ . [2017]

Column 1 Column 2 Column 3

(I) $W_{1 \to 2} = \frac{1}{γ - 1} (P_{2} V_{2} - P_{1} V_{1})$ (i) Isothermal (P) [IMAGE 482]

(II) $W_{1 \to 2} = - P V_{2} + P V_{1}$ (ii) Isochoric (Q) [IMAGE 483]

(III) $W_{1 \to 2} = 0$ (iii) Isobaric (R) [IMAGE 484]

(IV) $W_{1 \to 2} = - n R T \ln (\frac{V_{2}}{V_{1}})$ (iv) Adiabatic (S) [IMAGE 485]

Q. Which one of the following options correctly represents a thermodynamic process that is used as a correction in the determination of the speed of sound in an ideal gas?

(I)(ii)(Q)
(IV)(ii)(R)
(III)(iv)(R)
(I)(iv)(Q)

1

2

3

4

(4)

Laplace's correction of the speed of sound in ideal gas is related to adiabatic process.

P–V curve in adiabatic process is steeper than isothermal.

Q 26 :

By appropriately matching the information given in the three columns of the following table. An ideal gas is undergoing a cyclic thermodynamic process in different ways as shown in the corresponding P–V diagrams in column 3 of the table. Consider only the path from state 1 to state 2. W denotes the corresponding work done on the system. The equations and plots in the table have standard notations as used in thermodynamic processes. Here Y is the ratio of heat capacities at constant pressure and constant volume. The number of moles in the gas is $n$ . [2017]

Column 1 Column 2 Column 3

(I) $W_{1 \to 2} = \frac{1}{γ - 1} (P_{2} V_{2} - P_{1} V_{1})$ (i) Isothermal (P) [IMAGE 486]

(II) $W_{1 \to 2} = - P V_{2} + P V_{1}$ (ii) Isochoric (Q) [IMAGE 487]

(III) $W_{1 \to 2} = 0$ (iii) Isobaric (R) [IMAGE 488]

(IV) $W_{1 \to 2} = - n R T \ln (\frac{V_{2}}{V_{1}})$ (iv) Adiabatic (S) [IMAGE 489]

Q. Which one of the following options is the correct combination?

(IV)(ii)(S)
(III)(ii)(S)
(II)(iv)(P)
(II)(iv)(R)

1

2

3

4

(2)

Work done in isochoric process is zero for which we get a vertical line in P-V graph, graph(s).

Q 27 :

By appropriately matching the information given in the three columns of the following table. An ideal gas is undergoing a cyclic thermodynamic process in different ways as shown in the corresponding P–V diagrams in column 3 of the table. Consider only the path from state 1 to state 2. W denotes the corresponding work done on the system. The equations and plots in the table have standard notations as used in thermodynamic processes. Here Y is the ratio of heat capacities at constant pressure and constant volume. The number of moles in the gas is $n$ . [2017]

Column 1 Column 2 Column 3

(I) $W_{1 \to 2} = \frac{1}{γ - 1} (P_{2} V_{2} - P_{1} V_{1})$ (i) Isothermal (P) [IMAGE 490]

(II) $W_{1 \to 2} = - P V_{2} + P V_{1}$ (ii) Isochoric (Q) [IMAGE 491]

(III) $W_{1 \to 2} = 0$ (iii) Isobaric (R) [IMAGE 492]

(IV) $W_{1 \to 2} = - n R T \ln (\frac{V_{2}}{V_{1}})$ (iv) Adiabatic (S) [IMAGE 493]

Q. Which of the following options is the only correct representation of a process in which $Δ U = Δ Q - P Δ V$ ?

(II)(iv)(R)
(III)(iii)(P)
(II)(iii)(S)
(II)(iii)(P)

1

2

3

4

(4)

$Δ U = Δ Q - P Δ V$

$W = - P Δ V = - P (V_{2} - V_{1}) = - P V_{2} + P V_{1}$

$which is the formula for isobaric process, so graph (P)$

Q 28 :

One mole of a monatomic ideal gas is taken along two cyclic processes $E \to F \to G \to E$ and $E \to F \to H \to E$ as shown in the PV diagram. The processes involved are purely isochoric, isobaric, isothermal or adiabatic. [2013]

[IMAGE 494]

Match the paths in List I with the magnitudes of the work done in List II and select the correct answer using the codes given below the lists.

List I List II

P. $G \to E$ 1. $160 P_{0} V_{0} \ln 2$

Q. $G \to H$ 2. $36 P_{0} V_{0}$

R. $F \to H$ 3. $24 P_{0} V_{0}$

S. $F \to G$ 4. $31 P_{0} V_{0}$

Codes :

P-4, Q-3, R-2, S-1
P-4, Q-3, R-1, S-2
P-3, Q-1, R-2, S-4
P-1, Q-3, R-2, S-4

1

2

3

4

(1)

$W_{G E} = P_{0} (V_{0} - 32 V_{0}) = - 31 P_{0} V_{0}$

$W_{G H} = P_{0} (8 V_{0} - 32 V_{0}) = - 24 P_{0} V_{0}$

${(W_{F H})}_{adiabatic} = \frac{P_{f} V_{f} - P_{i} V_{i}}{1 - γ} = \frac{P_{0} (8 V_{0}) - 32 P_{0} (V_{0})}{1 - \frac{5}{3}} = 36 P_{0} V_{0}$

${(W_{F G})}_{isothermal} = n R T \ln (\frac{V_{f}}{V_{i}}) = P_{0} V_{0} \ln (\frac{V_{f}}{V_{i}})$

$= 1 (32 P_{0} V_{0}) \log_{e} (\frac{32 V_{0}}{V_{0}})$

$= 32 P_{0} V_{0} \log_{e} 2^{5} = 160 P_{0} V_{0} \log_{e} 2$

$(1) is the correct option$

Q 29 :

A thermally insulating cylinder has a thermally insulating and frictionless movable partition in the middle, as shown in the figure below. On each side of the partition, there is one mole of an ideal gas with specific heat at constant volume, $C_{v} = 2 R$ . Here, $R$ is the gas constant. Initially, each side has a volume $V_{0}$ and temperature $T_{0}$ . The left side has an electric heater, which is turned on at a very low power to transfer heat $Q$ to the gas on the left side. As a result the partition moves slowly towards the right reducing the right side volume to $V_{0} / 2$ . Consequently, the gas temperatures on the left and the right sides become $T_{L}$ and $T_{R}$ , respectively. Ignore the changes in the temperatures of the cylinder, heater and the partition.

[IMAGE 495]

Q. The value of $\frac{T_{R}}{T_{0}}$ is [2021]

$\sqrt{2}$
$\sqrt{3}$
$2$
$3$

1

2

3

4

(1)

[IMAGE 496]

$As C_{V} = 2 R \Rightarrow \frac{R}{γ - 1} = 2 R \Rightarrow γ = \frac{3}{2}$

$In adiabatic process, T V^{γ - 1} = constant$

$\Rightarrow T_{0} V_{0}^{1 / 2} = T_{R} {(\frac{V_{0}}{2})}^{1 / 2}$

$\Rightarrow T_{R} = \sqrt{2} T_{0}$

Q 30 :

A thermally insulating cylinder has a thermally insulating and frictionless movable partition in the middle, as shown in the figure below. On each side of the partition, there is one mole of an ideal gas with specific heat at constant volume, $C_{v} = 2 R$ . Here, $R$ is the gas constant. Initially, each side has a volume $V_{0}$ and temperature $T_{0}$ . The left side has an electric heater, which is turned on at a very low power to transfer heat $Q$ to the gas on the left side. As a result the partition moves slowly towards the right reducing the right side volume to $V_{0} / 2$ . Consequently, the gas temperatures on the left and the right sides become $T_{L}$ and $T_{R}$ , respectively. Ignore the changes in the temperatures of the cylinder, heater and the partition.

[IMAGE 497]

Q. The value of $\frac{Q}{R T_{0}}$ is [2021]

$4 (2 \sqrt{2} + 1)$
$4 (2 \sqrt{2} - 1)$
$(5 \sqrt{2} + 1)$
$(5 \sqrt{2} - 1)$

1

2

3

4

(2)

$∵$ $C_{V} = \frac{R}{γ - 1} = 2 R or γ - 1 = \frac{1}{2}$

$∴$ $γ = \frac{3}{2}$

$T_{0} V_{0}^{γ - 1} = T_{R} {(\frac{V_{0}}{2})}^{γ - 1}$ $[∵ Final volume of right side cylinder is reduced to \frac{V_{0}}{2}]$

$\Rightarrow \frac{T_{R}}{T_{0}} = \sqrt{2}$

$From P V^{γ} = constant$

$P {(\frac{V_{0}}{2})}^{γ} = P_{0} V_{0}^{γ} \Rightarrow P = P_{0} \times 2^{3 / 2}$

$\frac{P V}{T_{L}} = \frac{P_{0} V_{0}}{T_{0}} \Rightarrow T_{L} = 2^{3 / 2} \times \frac{3}{2} T_{0} = 3 \sqrt{2} T_{0}$

$Q = n C_{V} Δ T_{1} + n C_{V} Δ T_{2} = 2 R (T_{L} - T_{0}) + 2 R (T_{R} - T_{0})$ $[∵ n = 1]$

$= 2 R (3 \sqrt{2} - 1) T_{0} + 2 R (\sqrt{2} - 1) T_{0} = 4 R T_{0} (2 \sqrt{2} - 1)$

$∴$ $\frac{Q}{R T_{0}} = 2 (3 \sqrt{2} - 1) + 2 (\sqrt{2} - 1) = 8 \sqrt{2} - 4$

$= 4 (2 \sqrt{2} - 1)$

Topic Question Set

Q 2 :

An ideal gas is initially at $P_{1}, V_{1}$ is expanded to $P_{2}, V_{2}$ and then compressed adiabatically to the same volume $V_{1}$ and pressure $P_{3}$ . If $W$ is the net work done by the gas in complete process which of the following is true [2004]

Q 3 :

The PT diagram for an ideal gas is shown in the figure, where AC is an adiabatic process, find the corresponding PV diagram. [2003]

[IMAGE 455]

Q 4 :

An ideal gas is taken through the cycle A → B → C → A, as shown in the figure. If the net heat supplied to the gas in the cycle is 5 J, the work done by the gas in the process C → A is [2002]

[IMAGE 460]

Q 5 :

P-V plots for two gases during adiabatic processes are shown in the figure. Plots 1 and 2 should correspond respectively to [2001]

[IMAGE 461]

Q 9 :

A diatomic ideal gas is compressed adiabatically to $\frac{1}{32}$ of its initial volume. If the initial temperature of the gas is $T_{i}$ (in Kelvin) and the final temperature is $a T_{i}$ , the value of $a$ is ______. [2010]

Q 17 :

One mole of a monatomic ideal gas goes through a thermodynamic cycle, as shown in the volume versus temperature $(V - T)$ diagram. The correct statement(s) is/are: [R is the gas constant] [2019]

[IMAGE 473]

Q 18 :

One mole of an ideal gas in initial state $A$ undergoes a cyclic process $A B C A$ , as shown in the figure. Its pressure at $A$ is $P_{0}$ . Choose the correct option(s) from the following. [2010]

[IMAGE 475]

Q 19 :

The figure shows the $P - V$ plot of an ideal gas taken through a cycle $A B C D A$ . The part $A B C$ is a semi-circle and $C D A$ is half of an ellipse. Then, [2009]

[IMAGE 477]

Q 22 :

If the process carried out on one mole of monatomic ideal gas is shown in the figure $P V$ -diagram with $P_{0} V_{0} = \frac{1}{3} R T_{0},$ the correct match is, [2019]

[IMAGE 479]

Q 23 :

If the process carried out on one mole of monatomic ideal gas is as shown in the TV-diagram with $P_{0} V_{0} = \frac{1}{3} R T_{0},$ the correct match is, [2019]

[IMAGE 480]

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	List - I		List - II
(P)	Work done in the complete cyclic process	(1)	$R T_{0} - 4 R T_{0} \ln 2$
(Q)	Change in the internal energy of the gas in the process JK	(2)	$0$
(R)	Heat given to the gas in the process KL	(3)	$3 R T_{0}$
(S)	Change in the internal energy of the gas in the process MJ	(4)	$- 2 R T_{0} \ln 2$
		(5)	$- 3 R T_{0} \ln 2$

	List - I		List - II
(I)	$10^{- 3} kg$ of water at $100 ° C$ is converted to steam at the same temperature, at a pressure of $10^{5} Pa$ . The volume of the system changes from $10^{- 6} m^{3}$ to $10^{- 3} m^{3}$ in the process. Latent heat of water $= 2250 kJ/kg$ .	(P)	$2 kJ$
(II)	$0.2$ moles of a rigid diatomic ideal gas with volume $V$ at temperature $500 K$ undergoes an isobaric expansion to volume $3 V$ . Assume $R = 8.0 {J mol}^{- 1} K^{- 1}$ .	(Q)	$7 kJ$
(III)	One mole of a monatomic ideal gas is compressed adiabatically from volume $V = \frac{1}{3} m^{3}$ and pressure $2 kPa$ to volume $\frac{V}{8}$ .	(R)	$4 kJ$
(IV)	Three moles of a diatomic ideal gas whose molecules can vibrate, is given $9 kJ$ of heat and undergoes isobaric expansion.	(S)	$5 kJ$
		(T)	$3 kJ$

	LIST-I		LIST-II
P.	In process I	1.	Work done by the gas is zero
Q.	In process II	2.	Temperature of the gas remains unchanged
R.	In process III	3.	No heat is exchanged between the gas and its surroundings
S.	In process IV	4.	Work done by the gas is $6 P_{0} V_{0}$

	Column 1		Column 2		Column 3
(I)	$W_{1 \to 2} = \frac{1}{γ - 1} (P_{2} V_{2} - P_{1} V_{1})$	(i)	Isothermal	(P)	[IMAGE 482]
(II)	$W_{1 \to 2} = - P V_{2} + P V_{1}$	(ii)	Isochoric	(Q)	[IMAGE 483]
(III)	$W_{1 \to 2} = 0$	(iii)	Isobaric	(R)	[IMAGE 484]
(IV)	$W_{1 \to 2} = - n R T \ln (\frac{V_{2}}{V_{1}})$	(iv)	Adiabatic	(S)	[IMAGE 485]

	List I		List II
P.	$G \to E$	1.	$160 P_{0} V_{0} \ln 2$
Q.	$G \to H$	2.	$36 P_{0} V_{0}$
R.	$F \to H$	3.	$24 P_{0} V_{0}$
S.	$F \to G$	4.	$31 P_{0} V_{0}$

Topic Question Set

Q 2 : An ideal gas is initially at P1,V1 is expanded to P2,V2 and then compressed adiabatically to the same volume V1 and pressure P3. If W is the net work done by the gas in complete process which of the following is true [2004]

Q 3 : The PT diagram for an ideal gas is shown in the figure, where AC is an adiabatic process, find the corresponding PV diagram. [2003] [IMAGE 455]

Q 4 : An ideal gas is taken through the cycle A → B → C → A, as shown in the figure. If the net heat supplied to the gas in the cycle is 5 J, the work done by the gas in the process C → A is [2002] [IMAGE 460]

Q 5 : P-V plots for two gases during adiabatic processes are shown in the figure. Plots 1 and 2 should correspond respectively to [2001] [IMAGE 461]

Q 9 : A diatomic ideal gas is compressed adiabatically to 132 of its initial volume. If the initial temperature of the gas is Ti (in Kelvin) and the final temperature is aTi, the value of a is ______. [2010]

Q 15 : In the given P-V diagram, a monoatomic gas (γ=53) is first compressed adiabatically from state A to state B. Then it expands isothermally from state B to state C. [Given (13)0.6=0.5, ln2=0.7] [IMAGE 472] Which of the following statement(s) is(are) correct? [2022]

Q 17 : One mole of a monatomic ideal gas goes through a thermodynamic cycle, as shown in the volume versus temperature (V-T) diagram. The correct statement(s) is/are: [R is the gas constant] [2019] [IMAGE 473]

Q 18 : One mole of an ideal gas in initial state A undergoes a cyclic process ABCA, as shown in the figure. Its pressure at A is P0. Choose the correct option(s) from the following. [2010] [IMAGE 475]

Q 19 : The figure shows the P-V plot of an ideal gas taken through a cycle ABCDA. The part ABC is a semi-circle and CDA is half of an ellipse. Then, [2009] [IMAGE 477]

Q 22 : If the process carried out on one mole of monatomic ideal gas is shown in the figure PV-diagram with P0V0=13RT0, the correct match is, [2019] [IMAGE 479]

Q 23 : If the process carried out on one mole of monatomic ideal gas is as shown in the TV-diagram with P0V0=13RT0, the correct match is, [2019] [IMAGE 480]

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Q 2 :

An ideal gas is initially at $P_{1}, V_{1}$ is expanded to $P_{2}, V_{2}$ and then compressed adiabatically to the same volume $V_{1}$ and pressure $P_{3}$ . If $W$ is the net work done by the gas in complete process which of the following is true [2004]

Q 3 :

The PT diagram for an ideal gas is shown in the figure, where AC is an adiabatic process, find the corresponding PV diagram. [2003]

[IMAGE 455]

Q 4 :

An ideal gas is taken through the cycle A → B → C → A, as shown in the figure. If the net heat supplied to the gas in the cycle is 5 J, the work done by the gas in the process C → A is [2002]

[IMAGE 460]

Q 5 :

P-V plots for two gases during adiabatic processes are shown in the figure. Plots 1 and 2 should correspond respectively to [2001]

[IMAGE 461]

Q 9 :

A diatomic ideal gas is compressed adiabatically to $\frac{1}{32}$ of its initial volume. If the initial temperature of the gas is $T_{i}$ (in Kelvin) and the final temperature is $a T_{i}$ , the value of $a$ is ______. [2010]

Q 17 :

One mole of a monatomic ideal gas goes through a thermodynamic cycle, as shown in the volume versus temperature $(V - T)$ diagram. The correct statement(s) is/are: [R is the gas constant] [2019]

[IMAGE 473]

Q 18 :

One mole of an ideal gas in initial state $A$ undergoes a cyclic process $A B C A$ , as shown in the figure. Its pressure at $A$ is $P_{0}$ . Choose the correct option(s) from the following. [2010]

[IMAGE 475]

Q 19 :

The figure shows the $P - V$ plot of an ideal gas taken through a cycle $A B C D A$ . The part $A B C$ is a semi-circle and $C D A$ is half of an ellipse. Then, [2009]

[IMAGE 477]

Q 22 :

If the process carried out on one mole of monatomic ideal gas is shown in the figure $P V$ -diagram with $P_{0} V_{0} = \frac{1}{3} R T_{0},$ the correct match is, [2019]

[IMAGE 479]

Q 23 :

If the process carried out on one mole of monatomic ideal gas is as shown in the TV-diagram with $P_{0} V_{0} = \frac{1}{3} R T_{0},$ the correct match is, [2019]

[IMAGE 480]