First slide

Thermodynamic processes

Question

A mono atomic gas at a pressure P, having a volume V, expands isothermally to a volume 2V, and then adiabatically to a volume 16 V. The final pressure of the gas is: $(δ = \frac{5}{3})$

Moderate

A

64 P
B

32 P
C

$\frac{P}{64}$
D

16 P

Solution

1st Isothermal process
${given initial pressure P}_{1} {=P , initial volume V}_{1} =V f i n a l p r e s s u r e b e P_{2} {then volume is V}_{2} =2V$
$i s o t h e r m a l p r o c e s s, r e l a t i o n b e t w e e n P r e s s u r e P, a n d v o l u m e V i s$
$Pα \frac{1}{V}$
$\Rightarrow \frac{P_{1}}{P_{2}} = \frac{V_{2}}{V_{1}} ---(1)$
$s u b s t i t u t e g i v e n v a l u e s i n (1) e q n \frac{P}{P_{2}} = \frac{2 V}{V}$
$\Rightarrow P_{2} =P/2$
2nd adiabatic process
$\begin{array}{rcl} {given final volume of the gas is V}_{3} =16V \end{array}$
$\begin{array}{rcl} I n a d i a b a t i c p r o c e s s r e l a t i o n b e t w e e n p r e s s u r e P a n d v o l u m e V i s P_{2} {V_{2}}^{δ} {=P}_{3} {V_{3}}^{δ} \end{array}$
$\begin{array}{rcl} g i v e n s p e c i f i c h e a t r a t i o δ & = & \frac{5}{3} \end{array}$
$\begin{array}{rcl} \frac{P_{2}}{P_{3}} = {(\frac{V_{3}}{V_{2}})}^{5/3} \end{array}$
$\begin{array}{l} s u b s t i t u t e t h e g i v e n a n d o b t a i n e d v a l u e s, l e g i b l y \\ \frac{\frac{P}{2}}{P_{3}} = {(\frac{16V}{2V})}^{5/3} \end{array}$
$\begin{array}{l} \frac{P}{2 P_{3}} {=2}^{5} \end{array}$
$f i n a l p r e s s u r e P_{3} = \frac{P}{64}$

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