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 , pressure and volume and the spring is in its relaxed state. The gas is then heated very slowly to temp

Question

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]

Smart Achievers · Accepted Answer

(1, 2, 3)

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

$(1) As gas is ideal monoatomic,$

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

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

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

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

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

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

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

Solution

1 If $V_{2} = 2 V_{1}$ and $T_{2} = 3 T_{1}$ , then the energy stored in the spring is $\frac{1}{4} P_{1} V_{1}$

2 If $V_{2} = 2 V_{1}$ and $T_{2} = 3 T_{1}$ , then the change in internal energy is $3 P_{1} V_{1}$

3 If $V_{2} = 3 V_{1}$ and $T_{2} = 4 T_{1}$ , then the work done by the gas is $\frac{7}{3} P_{1} V_{1}$

4 If $V_{2} = 3 V_{1}$ and $T_{2} = 4 T_{1}$ , then the heat supplied to the gas is $\frac{17}{6} P_{1} V_{1}$

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Solution

1 If V2=2V1 and T2=3T1, then the energy stored in the spring is 14P1V1

2 If V2=2V1 and T2=3T1, then the change in internal energy is 3P1V1

3 If V2=3V1 and T2=4T1, then the work done by the gas is 73P1V1

4 If V2=3V1 and T2=4T1, then the heat supplied to the gas is 176P1V1

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1 If $V_{2} = 2 V_{1}$ and $T_{2} = 3 T_{1}$ , then the energy stored in the spring is $\frac{1}{4} P_{1} V_{1}$

2 If $V_{2} = 2 V_{1}$ and $T_{2} = 3 T_{1}$ , then the change in internal energy is $3 P_{1} V_{1}$

3 If $V_{2} = 3 V_{1}$ and $T_{2} = 4 T_{1}$ , then the work done by the gas is $\frac{7}{3} P_{1} V_{1}$

4 If $V_{2} = 3 V_{1}$ and $T_{2} = 4 T_{1}$ , then the heat supplied to the gas is $\frac{17}{6} P_{1} V_{1}$