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.

Question

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.	(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.	(s)	The gas gains heat

Smart Achievers · Accepted Answer

(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.

Solution

1 A → p, s; B → p, r; C → q; D → q, s

2 A → q; B → p, r; C → p, s; D → q, s

3 A → p, s; B → q, s; C → q; D → p, r

4 A → q, s, s; B → p; C → q; D → p, r

Want Us to Email you About Special Offers & Updates?