JEE Advanced First Law of Thermodynamics PYQ with Solutions

Q 1 :

In a given process on an ideal gas, $d W = 0$ and $d Q < 0$ . Then for the gas [2001]

the temperature will decrease
the volume will increase
the pressure will remain constant
the temperature will increase

1

2

3

4

(1)

$From the first law of thermodynamics$

$d Q = d U + d W$

Here $d W = 0 (given)$

$∴$ $d Q = d U$

Now since $d Q < 0 (given)$

i.e., $d Q$ is negative so $d U$ decreases.

Internal energy $' U'$ decreases when temperature $T$ decreases.

Q 2 :

A thermodynamic system is taken from an initial state $i$ with internal energy $U_{i} = 100 J$ to the final state $f$ along two different paths $i a f$ and $i b f$ , as schematically shown in the figure. The work done by the system along the paths $a f$ , $i b$ and $b f$ are $W_{a f} = 200 J,$ $W_{i b} = 50 J$ and $W_{b f} = 100 J$ respectively. The heat supplied to the system along the path $i a f$ , $i b$ and $b f$ are $Q_{i a f}$ , $Q_{i b}$ and $Q_{b f}$ respectively. If the internal energy of the system in the state $b$ is $U_{b} = 200 J$ and $Q_{i a f} = 500 J,$ then the ratio $\frac{Q_{b f}}{Q_{i b}}$ is [2014]

[IMAGE 446]

(2)

Applying first law of thermodynamics to path $i a f$

$Q_{i a f} = Δ U_{i a f} + W_{i a f}$

$500 = Δ U_{i a f} + 200$

$∴$ $Δ U_{i a f} = 300 J$

Now,

$Q_{i b f} = Δ U_{i b f} + W_{i b} + W_{b f}$

$Q_{i b} + Q_{b f} = 300 + 50 + 100 = 450 J ...(i)$

Also $Q_{i b} = Δ U_{i b} + W_{i b}$

$∴$ $Q_{i b} = 100 + 50 = 150 J ...(ii)$

From eq. (i) & (ii) $\frac{Q_{b f}}{Q_{i b}} = \frac{Q_{i b f} - Q_{i b}}{Q_{i b}} = \frac{300}{150} = 2$

Q 3 :

An ideal monoatomic gas is confined in a horizontal cylinder by a spring loaded piston (as shown in the figure). Initially the gas is at temperature $T_{1}$ , pressure $P_{1}$ and volume $V_{1}$ and the spring is in its relaxed state. The gas is then heated very slowly to temperature $T_{2}$ , pressure $P_{2}$ and volume $V_{2}$ . During this process the piston moves out by a distance $x$ .

Ignoring the friction between the piston and the cylinder, the correct statement(s) is/are [2015]

[IMAGE 447]

If $V_{2} = 2 V_{1}$ and $T_{2} = 3 T_{1}$ , then the energy stored in the spring is $\frac{1}{4} P_{1} V_{1}$
If $V_{2} = 2 V_{1}$ and $T_{2} = 3 T_{1}$ , then the change in internal energy is $3 P_{1} V_{1}$
If $V_{2} = 3 V_{1}$ and $T_{2} = 4 T_{1}$ , then the work done by the gas is $\frac{7}{3} P_{1} V_{1}$
If $V_{2} = 3 V_{1}$ and $T_{2} = 4 T_{1}$ , then the heat supplied to the gas is $\frac{17}{6} P_{1} V_{1}$

1

2

3

4

Select one or more options

(1, 2, 3)

Let spring is compressed by $x$ on heating the gas.

[IMAGE 448]

$(1) As gas is ideal monoatomic,$

$∴$ $\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}} ...(i)$

Force on spring by gas $= k x$

$∴$ $P_{2} = P_{1} + \frac{k x}{A} (A = area of cross-section of piston) ...(ii)$

When $V_{2} = 2 V_{1}, T_{2} = 3 T_{1}$

$∴$ $\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} (2 V_{1})}{3 T_{1}}$ $\Rightarrow P_{2} = \frac{3}{2} P_{1}$

Putting this value of $P_{2}$ in eqn. (ii) we get

$\frac{3}{2} P_{1} = P_{1} + \frac{k x}{A}$ $\Rightarrow k x = \frac{P_{1} A}{2}$

$x = \frac{V_{2} - V_{1}}{A} = \frac{2 V_{1} - V_{1}}{A} = \frac{V_{1}}{A}$

Energy stored in the spring $= \frac{1}{2} k x^{2} = \frac{1}{2} (k x) (x) = \frac{P_{1} V_{1}}{4}$

$(2) Change in internal energy,$

$Δ U = \frac{f}{2} (P_{2} V_{2} - P_{1} V_{1})$ $= \frac{3}{2} (\frac{3}{2} P_{1} \times 2 V_{1} - P_{1} V_{1}) = 3 P_{1} V_{1}$

$(3)$ Again, when $V_{2} = 3 V_{1}$ and $T_{2} = 4 T_{1}$ then

From eqn. (i),

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} (3 V_{1})}{4 T_{1}}$ $\Rightarrow P_{2} = \frac{4}{3} P_{1}$ $\Rightarrow x = \frac{V_{2} - V_{1}}{A} = \frac{2 V_{1}}{A}$

From eqn. (ii),

$\frac{4}{3} P_{1} = P_{1} + \frac{k x}{A}$ $\Rightarrow k x = \frac{P_{1} A}{3}$

Work done by gas = Work done by gas on atmosphere + Energy stored in spring

$W_{g} = P_{1} A x + \frac{1}{2} k x^{2}$ $= P_{1} (2 V_{1}) + \frac{1}{2} (\frac{P_{1} A}{3}) (\frac{2 V_{1}}{A})$

$= 2 P_{1} V_{1} + \frac{1}{3} P_{1} V_{1} = \frac{7}{3} P_{1} V_{1}$

$(4)$ $Δ Q = W_{g} + Δ U$ $= \frac{7}{3} P_{1} V_{1} + \frac{3}{2} (P_{2} V_{2} - P_{1} V_{1})$

$= \frac{7}{3} P_{1} V_{1} + \frac{3}{2} (\frac{4}{3} P_{1} \times 3 V_{1} - P_{1} V_{1})$

$= \frac{7}{3} P_{1} V_{1} + 6 P_{1} V_{1} - \frac{3}{2} P_{1} V_{1}$ $= P_{1} V_{1} (\frac{14 + 36 - 9}{6}) = \frac{41}{6} P_{1} V_{1}$

Q 4 :

One mole of a monoatomic gas is taken through a cycle ABCDA as shown in the P-V diagram. Column II gives the characteristics involved in the cycle. Match them with each of the processes given in Column I. [2011]

[IMAGE 449]

Column I Column II

(A) Process A → B (p) Internal energy decreases

(B) Process B → C (q) Internal energy increases

(C) Process C → D (r) Heat is lost

(D) Process D → A (s) Heat is gained

(t) Work is done on the gas

$A - p, r, t; B - p, r; C - q, s; D - r, t$
$A - r, t; B - p, r; C - q, s; D - p, r, t$
$A - r, t; B - q, s; C - p, r; D - p, r, t$
$A - r, t; B - p, r, t; C - p, r; D - q, s$

1

2

3

4

(1)

(1) Process $A \to B$

This is an isobaric process, P = constant and volume $(V)$ of the gas decreases. Therefore work is done on the gas.

$W = P (3 V - V) = 2 P V$

Also $V$ decreases so temperature at $B$ decreases.

$∴$ Internal energy U decreases.

From, $Q = U + W$ as $U$ and $W$ decrease so $Q$ decreases, that means heat is lost.

(2) Process $B \to C$

This is an isochoric process, V = constant pressure decreases,

$P \propto T$ so temperature also decreases.

$W = 0; Δ U = negative$

so $Δ Q negative$

Hence heat is lost.

(3) Process $C \to D$

This is isobaric, Pressure P = constant $V$ increases and $V \propto T$

so $T$ increases.

Hence $Δ W, Δ U and Δ Q$ are positive, so heat is gained by the gas.

(4) Process $D \to A$

Applying $P V = n R T$

for $D$ $P (9 V) = 1 R T_{D}$ $∴$ $T_{D} = \frac{9 P V}{R}$

for $A$ $3 P (3 V) = 1 R T_{A}$ $∴$ $T_{A} = \frac{9 P V}{R}$

i.e., the process is isothermal. $∴$ $Δ U = 0$

Now, $Δ Q = Δ U + W$ $∴$ $Δ Q = W$

As volume decreases in this process so $W$ is negative i.e., work done on the gas and $Δ Q$ is negative, hence heat is lost.

Q 5 :

Column I contains a list of processes involving expansion of an ideal gas. Match this with Column II describing the thermodynamic change during this process. Indicate your answer by darkening the appropriate bubbles of the 4 × 4 matrix given in the ORS. [2008]

Column I Column II

(A)
An insulated container has two chambers separated by a valve. Chamber I contains an ideal gas and Chamber II has vacuum. The valve is opened.

[IMAGE 450]
(p) The temperature of the gas decreases

(B) An ideal monoatomic gas expands to twice its remains original volume such that its pressure $P \propto 1 / V^{2}$ where V is the volume of the gas. (q) The temperature of the gas increases or constant

(C) An ideal monoatomic gas expands to twice its original volume such that its pressure $P \propto 1 / V^{4 / 3}$ (r) The gas loses heat where V is its volume

(D)
An ideal monoatomic gas expands such that pressure P and volume V follows the behaviour shown in the graph.

[IMAGE 451]
(s) The gas gains heat

A → p, s; B → p, r; C → q; D → q, s
A → q; B → p, r; C → p, s; D → q, s
A → p, s; B → q, s; C → q; D → p, r
A → q, s, s; B → p; C → q; D → p, r

1

2

3

4

(2)

$(A)-(q):$ As the ideal gas expands in vacuum, and the container is insulated therefore $W = 0 & Q = 0$ and according to first law of thermodynamics

$Δ U = Q + W$ $\Rightarrow Δ U = 0$

Hence there is no change in the temperature of the gas of $T$ is constant.

$(B)-(p,r):$ Given $P \propto \frac{1}{V^{2}}$ or, $P V^{2} = constant$

or, $n R T V = constant$ $∴$ $V \times T = constant$

As the gas expands its volume increases so temperature decreases.

We know that $Q = n C Δ T ...(i)$

For a polytropic process

$C = C_{v} + \frac{R}{1 - n}$ and $P V^{n} = constant$

Here $P V^{2} = constant$ $∴$ $n = 2$

$∴$ $C = C_{v} + \frac{R}{1 - 2} = C_{v} - R = \frac{3}{2} R - R = \frac{R}{2}$

$Q = n C Δ T = n \times \frac{R}{2} \times Δ T$

i.e., $Δ T$ is negative, $Q$ is negative so heat is lost.

As volume increases, so temperature decreases given

$P \propto \frac{1}{V^{4 / 3}}$ $\Rightarrow P V^{4 / 3} = constant$

$∴$ $n = \frac{4}{3}$

$∴$ $C = C_{v} + \frac{R}{1 - \frac{4}{3}} = \frac{3}{2} R + \frac{3 R}{- 1} = \frac{3}{2} R - 3 R = - \frac{3 R}{2}$

$∴$ $Q = n C_{v} Δ T = n (- \frac{3 R}{2}) Δ t$

As $Δ T$ is negative, $Q$ is positive. So gas gains heat.

$(D)-(q,s):$ From $P V = n R T$ $\Rightarrow T = \frac{P V}{n R}$

$P V$ increases so $T$ increases, volume increases so $W$ increases.

From $Q = Δ U + W,$ $Q$ increases.

Hence the gas gains heat.

Q 6 :

Heat given to process is positive, match the following option of Column I with the corresponding option of Column II : [2006]

[IMAGE 452]

Column I Column II

(A) JK (p) $Δ W > 0$

(B) KL (q) $Δ Q < 0$

(C) LM (r) $Δ W < 0$

(D) MJ (s) $Δ Q > 0$

A-p, B-q, C-r, D-s
A-q, B-p, C-s, D-r
A-r, B-s, C-p, D-q
A-s, B-r, C-q, D-p

1

2

3

4

(2)

From the given $P - V$ graph, in process $J \to K$ volume,

$V =$ constant $p$ is decreasing and $P \propto T$

Therefore, $T$ should also decrease.

$∴$ $W = P d V = 0, Δ V = Δ Q < 0 (negative)$

From $Δ Q = Δ U + Δ W$

In process $K \to L$ $P$ = constant $= n C_{p} Δ T$ $V$ is increasing so temperature should also increase.

$∴$ $Δ W = P d V > 0, Δ U = n C_{v} Δ T > 0$ and $Q = m C Δ T > 0$

In process $L \to M$ $V$ = constant $P$ increases to $T$ increases.

$∴$ $W = 0, Δ U > 0$ and $Q > 0$

In process $M \to J$

$V$ is decreasing $∴$ $Δ W < 0$

$T$ is also decreasing $∴$ $Δ U < 0$ and $Δ Q < 0$

Topic Question Set

Q 1 :

In a given process on an ideal gas, $d W = 0$ and $d Q < 0$ . Then for the gas [2001]

Q 6 :

Heat given to process is positive, match the following option of Column I with the corresponding option of Column II : [2006]

[IMAGE 452]

Column I Column II

(A) JK (p) $Δ W > 0$

(B) KL (q) $Δ Q < 0$

(C) LM (r) $Δ W < 0$

(D) MJ (s) $Δ Q > 0$

Want Us to Email you About Special Offers & Updates?

	Column I		Column II
(A)	Process A → B	(p)	Internal energy decreases
(B)	Process B → C	(q)	Internal energy increases
(C)	Process C → D	(r)	Heat is lost
(D)	Process D → A	(s)	Heat is gained
		(t)	Work is done on the gas

	Column I		Column II
(A)	An insulated container has two chambers separated by a valve. Chamber I contains an ideal gas and Chamber II has vacuum. The valve is opened. [IMAGE 450]	(p)	The temperature of the gas decreases
(B)	An ideal monoatomic gas expands to twice its remains original volume such that its pressure $P \propto 1 / V^{2}$ where V is the volume of the gas.	(q)	The temperature of the gas increases or constant
(C)	An ideal monoatomic gas expands to twice its original volume such that its pressure $P \propto 1 / V^{4 / 3}$	(r)	The gas loses heat where V is its volume
(D)	An ideal monoatomic gas expands such that pressure P and volume V follows the behaviour shown in the graph. [IMAGE 451]	(s)	The gas gains heat

	Column I		Column II
(A)	JK	(p)	$Δ W > 0$
(B)	KL	(q)	$Δ Q < 0$
(C)	LM	(r)	$Δ W < 0$
(D)	MJ	(s)	$Δ Q > 0$

Topic Question Set

Q 1 : In a given process on an ideal gas, dW=0 and dQ<0. Then for the gas [2001]

Q 6 : Heat given to process is positive, match the following option of Column I with the corresponding option of Column II : [2006] [IMAGE 452] Column I Column II (A) JK (p) ΔW>0 (B) KL (q) ΔQ<0 (C) LM (r) ΔW<0 (D) MJ (s) ΔQ>0

Want Us to Email you About Special Offers & Updates?

Q 1 :

In a given process on an ideal gas, $d W = 0$ and $d Q < 0$ . Then for the gas [2001]

Q 6 :

Heat given to process is positive, match the following option of Column I with the corresponding option of Column II : [2006]

[IMAGE 452]

Column I Column II

(A) JK (p) $Δ W > 0$

(B) KL (q) $Δ Q < 0$

(C) LM (r) $Δ W < 0$

(D) MJ (s) $Δ Q > 0$