Question

The initial pressure & volume of a given mass of gas with $\frac{C_P}{C_V}=\frac{7}{5}$ are $2\text{atm}$ & $4.8{\text{m}}^3$. The gas can exchange heat with the surroundings. It is suddenly compressed to $1.2{\text{m}}^3$ and then slowly compressed to $0.6{\text{m}}^3$. Find the final pressure of the gas.

• A. $2^{\frac{11}{4}}\text{atm}$
• B. $4\text{atm}$
• C. $16\text{atm}$
• D. $2^{\frac{24}{5}}\text{atm}$

Answer 1

The correct option is D $2^{\frac{24}{5}}\text{atm}$

The entire process can be represented as
$(\frac{P_1=2\text{atm}}{V_1=4.8{\text{m}}^3})\underset{\left[\text{Adiabatic}\right]}{\underset{\text{compressed}}{\overset{\text{Suddenly}}{\to }}}(\frac{P_2=P\text{atm}}{V_2=1.2m^3})\underset{\left[\text{Isothermal}\right]}{\underset{\text{compressed}}{\overset{\text{Slowly}}{\to }}}(\frac{P_3=P^{\prime }\text{atm}}{V_3=0.6{\text{m}}^3})$
From the equation of state for an adiabatic process in terms of $P$ and $V$ we can write that,
$P_1{V}_1^{\gamma }=P_2{V}_2^{\gamma }$
from the data given in the question,
$2\times (4.8{)}^{\frac{7}{5}}=P\times (1.2{)}^{\frac{7}{5}}$
$\Rightarrow P=2\times (4{)}^{\frac{7}{5}}.........(1)$
We know that equation of state for an isothermal process is given by,
$P_2{V}_2=P_3{V}_3$
$P\times 1.2=P^{\prime }\times 0.6$
$\Rightarrow P^{\prime }=2P$
$P^{\prime }=2\times 2\times 4^{\frac{7}{5}}=2^{\frac{24}{5}}\text{atm}$
Thus, option (d) is the correct answer.