An ideal gas undergoes a process maintaining relation between pressure (P) and volume (V) as where and are constants

Question

An ideal gas undergoes a process maintaining relation between pressure (P) and volume (V) as $P = P_{0} (1 + {{(\frac{V_{0}}{V})}^{2})}^{- 1},$ where $P_{0}$ and $V_{0}$ are constants.

If two samples A and B (two moles each) with initial volume $V_{0}$ and $3 V_{0}$ respectively undergo above mentioned process and attain same pressure, then the difference at the temperatures of these samples, $T_{B} - T_{A}$ is _______. (R = gas constant) [2026]

Smart Achievers · Accepted Answer

(2)

$P V = n R T \Rightarrow T = \frac{P V}{n R}$

$T = \frac{P_{0}}{n R} {[1 + {(\frac{V_{0}}{V})}^{2}]}^{- 1} V =$ $\frac{P_{0} V}{n R (1 + {(\frac{V_{0}}{V})}^{2})}$

$For sample A: T_{A} = \frac{P_{0} V_{0}}{2 R \cdot 2} = \frac{P_{0} V_{0}}{4 R}$

$For sample B: T_{B} = \frac{P_{0} 3 V_{0}}{2 R [1 + \frac{1}{9}]} = \frac{27 P_{0} V_{0}}{20 R}$

$T_{B} - T_{A} = \frac{P_{0} V_{0}}{R} [\frac{27}{20} - \frac{1}{4}] = \frac{11 P_{0} V_{0}}{10 R}$

Solution

1 $\frac{9 P_{0} V_{0}}{8 R}$

2 $\frac{11 P_{0} V_{0}}{10 R}$

3 $\frac{7 P_{0} V_{0}}{6 R}$

4 $\frac{13 P_{0} V_{0}}{11 R}$

Want Us to Email you About Special Offers & Updates?

Solution

1 9P0V08R

2 11P0V010R

3 7P0V06R

4 13P0V011R

Ans. (2) PV=nRT⇒T=PVnR T=P0nR[1+(V0V)2]-1V=P0VnR(1+(V0V)2) For sample A: TA=P0V02R·2=P0V04R For sample B: TB=P0 3V02R[1+19]=27P0V020R TB-TA=P0V0R[2720-14]=11P0V010R

Want Us to Email you About Special Offers & Updates?

1 $\frac{9 P_{0} V_{0}}{8 R}$

2 $\frac{11 P_{0} V_{0}}{10 R}$

3 $\frac{7 P_{0} V_{0}}{6 R}$

4 $\frac{13 P_{0} V_{0}}{11 R}$