Question

A monoatomic gas at a pressure P, having a volume V expands isothermally to a volume 2V and then adiabatically to a volume 16V. The final pressure of the gas is: (Take $\gamma= \dfrac{5}{3}$)
A. 64P
B. 32P
C. $\dfrac {P}{64}$
D. 16P

Answer 1

Hint: To solve this problem, use the formula for Boyle’s law to find the intermediate pressure in terms of initial temperature. Then use the relationship between pressure and volume for an adiabatic system. Substitute the values of gamma, intermediate pressure, and final pressure & intermediate volume and final volume. Solve this obtained equation and find the value of ${P}_{3}$ which is the final pressure of a gas.
Formula used:
${P}_{1}{V}_{1}= {P}_{2}{V}_{2}$
${P}_{2}{V}_{2}^{\gamma}={P}_{3}{V}_{3}^{\gamma}$

Complete step-by-step solution:
Given: Initial Volume ${V}_{1}$= V
Intermediate Volume ${V}_{2}$= 2V
Final Volume ${V}_{3}$= 16V
Initial Pressure ${P}_{1}$= P
Intermediate Pressure= ${P}_{2}$
$\gamma = \dfrac {5}{3}$
Let the final pressure of the gas be ${P}_{3}$.
According to Boyle’s Law,
${P}_{1}{V}_{1}= {P}_{2}{V}_{2}$
Substituting the values in above equation we get,
$PV= {P}_{2}2V$
$\Rightarrow {P}_{2}= \dfrac {P}{2}$
We also know, adiabatically, the relationship between pressure and volume is given by,
${P}_{2}{V}_{2}^{\gamma}={P}_{3}{V}_{3}^{\gamma}$
Substituting values in above equation we get,
$\dfrac {P}{2}{(2V)}^{\dfrac {5}{3}}= {P}_{3}{(16V)}^{\dfrac {5}{3}}$
$\Rightarrow \dfrac {P}{2}{(2)}^{\dfrac {5}{3}}={P}_{3}{(16)}^{\dfrac {5}{3}}$
$\Rightarrow {P}_{3}= \dfrac {P}{2}{(\dfrac {2}{16})}^{\dfrac {5}{3}}$
$\Rightarrow {P}_{3}=\dfrac {P}{2}{(\dfrac {1}{8})}^{\dfrac {5}{3}}$
$\Rightarrow {P}_{3}=\dfrac {P}{2} {(\dfrac {1}{8})}^{\dfrac {5}{3}}$
$\Rightarrow {P}_{3}=\dfrac {P}{2} {(\dfrac {1}{{2}^{3}})}^{\dfrac {5}{3}}$
$\Rightarrow {P}_{3}=\dfrac {P}{2} (\dfrac {1}{{2}^{5}})$
$\Rightarrow {P}_{3}=\dfrac {P}{{2}^{6}}$
$\Rightarrow {P}_{3}= \dfrac {P}{64}$
Hence, the final pressure of the gas is $\dfrac {P}{64}$.
So, the correct answer is option C i.e. $\dfrac {P}{64}$.

Note: $\gamma$ is the specific heat at constant pressure to that at a constant volume. Here, the value of $\gamma$ for monoatomic gas is given. While sometimes, it is not mentioned. So, students must remember the value of $\gamma$ for at least monoatomic, diatomic and triatomic molecules. For monoatomic gas $\gamma= \dfrac {5}{3}$, for diatomic gas $\gamma= \dfrac {7}{5}$ and for triatomic molecules $\gamma= \dfrac {9}{7}$.