Home KINETIC THEORY OF GASES Previous year Questions THERMODYNAMICS A monoatomic gas at pressure P, having volume V expands isothermally to a volume 2V &then adiabatically to volume 16V. The final pressure of gas is ..( take gamma =5/3)

A monoatomic gas at pressure P, having volume V expands isothermally to a volume 2V &then adiabatically to volume 16V. The final pressure of gas is ..( take gamma =5/3)

July 24, 2021 Add Comment KINETIC THEORY OF GASES , Previous year Questions , THERMODYNAMICS Edit

[AIPMT 2014]

a. 64 P b. 32 P c. P/64 d. 16 P

PHYSICS WITH AKSHAY GOLE

Solution:

In osothermal process the temperature of system remains constant

Therefore ideal gas equation becomes,

$PV=constant$

$P_1V_1=P_2V_2$

Here,

$V_2=2V_1$

- - - - (Given)

Putting values

$P_1V_1=P_2(2V_1)$

$P_2=\frac{P_1}{2}$

In an adiabatic equation

$PV^{\gamma}=constant$

$P_2 V_2^{\gamma}=P_3 V_3^{\gamma}$

$\frac{P_1}{2}\; 2^{\gamma } V_1^{\gamma}=P_3\; 16^{\gamma} V_1^{\gamma}$

$P_3=\frac{P_1}{2}\;\left ( \frac{2}{16} \right )^{\gamma }$

$P_3 = \frac{P_1}{2} \left ( \frac{2}{16} \right )^\gamma$

but,

$\gamma = \frac{5}{3}$

$P_3=\frac{P_1}{2}\;\left ( \frac{2}{16} \right )^{\frac{5}{3} }$

$P_3=\frac{P_1}{2}\;\left ( \frac{1}{8} \right )^{\frac{5}{3} }$

$P_3=\frac{P_1}{2}\;\left ( \frac{1}{8^{\frac{5}{3}} } \right )$

$P_3=\frac{P_1}{2}\;\left ( \frac{1}{32 } \right )$

$P_3=\frac{P_1}{64}$

Therefore, the correct answer is option c: P/64

Milan Tomic

Hi. I’m Designer of Blog Magic. I’m CEO/Founder of ThemeXpose. I’m Creative Art Director, Web Designer, UI/UX Designer, Interaction Designer, Industrial Designer, Web Developer, Business Enthusiast, StartUp Enthusiast, Speaker, Writer and Photographer. Inspired to make things looks better.

0 comments:

Post a Comment

Subscribe to: Post Comments ( Atom )