Question

The initial pressure & volume of a given mass of gas $(\frac{C_P}{C_V}=\frac{5}{3})$ are $P_0$ and $V_0$. The gas can exchange heat with the surroundings. It is slowly compressed to volume $\frac{V_0}{4}$ and then suddenly compressed to volume $\frac{V_0}{8}$ and then again slowly compressed to $\frac{V_0}{16}$. Find the final pressure.

• A. $2^{\frac{14}{3}}{P}_0$
• B. $2^{\frac{11}{3}}{P}_0$
• C. $2^{\frac{7}{3}}{P}_0$
• D. $2^{\frac{5}{3}}{P}_0$

Answer 1

The correct option is A $2^{\frac{14}{3}}{P}_0$

We know that, when a gas is compressed slowly, the process is isothermic and when it's quickly compressed, process is adiabatic.

Step I (isothermal compression) :
We know that equation of state for an isothermal process is given by,
$P_1{V}_1=P_2{V}_2$
From the data given in the question,
$P_0\times V_0=P^{\prime }\times \frac{V_0}{4}$
$\Rightarrow P^{\prime }=4P_0......\left(1\right)$

Step II (adiabatic compression) :
From the equation of state for an adibatic process in terms of $P$ and $V$, we can write that
$P_2{V}_2^{\gamma }=P_3{V}_3^{\gamma }$
From the data given in the question,
$P^{\prime }\times {\left(\frac{V_0}{4}\right)}^{\frac{5}{3}}=P^{\prime \prime }\times {\left(\frac{V_0}{8}\right)}^{\frac{5}{3}}$
Using $\left(1\right)$,
$P^{\prime \prime }=4P_0\times (2{)}^{\frac{5}{3}}......(2)$
$\Rightarrow P^{\prime \prime }=P_0\times (2{)}^{\frac{5}{3}+2}=2^{\frac{11}{3}}{P}_0\text{atm}$

Step III (isothermal compression) :
We can write from the data given in the question that,
$P^{\prime \prime }\frac{V_0}{8}=P^{\prime \prime \prime }\frac{V_0}{16}$
$\Rightarrow P^{\prime \prime \prime }=2\times 2^{\frac{11}{3}}{P}_0$
$\Rightarrow P^{\prime \prime \prime }=2^{\frac{14}{3}}{P}_0$
Thus, option (a) is the correct answer.