You visited us 1 times! Enjoying our articles? Unlock Full Access!

Ideal Gas Equation from KTG

One mole of a...

Question

One mole of an ideal gas at initial temperature $T$ , undergoes a quasi -static process, during which the volume $V$ is doubled. During the process, the internal energy $U$ varies with volume as $U = a V^{3}$ , where $a$ is a constant. The work done during this process is $p R T$ where $p$ is

Open in App

Solution

Given
Internal energy $U = a V^{3} \Rightarrow \frac{f n R T}{2} = a V^{3}$
$n R T = \frac{2 a V^{3}}{f} . . . . . . . . . . (1)$
where $f$ number of degrees of freedom.

Ideal gas equation $P V = n R T$ $\Rightarrow P V = (\frac{2 a}{f}) V^{3} \Rightarrow P = (\frac{2 a}{f}) V^{2}$
Now, work done when volume is doubled is $W = V_{2} \int V_{1} P d V = 2 V \int V (\frac{2 a}{f}) V^{2} d V = (\frac{2 a}{f}) \frac{(8 - 1)}{3} V^{3} = (\frac{2 a}{f}) \frac{7 V^{3}}{3} = \frac{7 n R T}{3} = \frac{7 R T}{3}$ (as $n = 1$ & from $(1)$ )

OR

Given Internal energy $U \propto V^{3}$ and $U \propto T$
$\Rightarrow T \propto V^{3} . . . . . . . . . . . (1)$
If $V$ is doubled, $T$ increases $8$ times. So, increase in temperature $Δ T = 8 T - T = 7 T$
From ideal gas equation $T \propto P V$
From $(1)$
$\Rightarrow P V \propto V^{3}$
$\Rightarrow P \propto V^{2}$
$\Rightarrow P V^{- 2} = C o n s t a n t$
Its a Polytropic process with polytropic coefficient $x = - 2$
So Work done in polytropic process $W = \frac{n R Δ T}{1 - x}$
$W = \frac{(1) R (7 T)}{1 - (- 2)}$
$W = \frac{7 R T}{3}$

Suggest Corrections

Similar questions

U = U_{0} + 2 P V

P

V

U_{0}

A B C D

T = - α V^{3} + β V^{2}

α

β

P - V

1 \to 2

(300 K)

2 \to 3

40 J

3 \to 1

T_{3}

275 K

3 \to 1

V

P

Join BYJU'S Learning Program