Question

One mole of a monoatomic gas behaving as per $PV=nRT$ at $27 \, ^\circ C$ is subjected to reversible adiabatic compression until the final temperature reaches $327 \, ^\circ C$ . If the initial pressure was $1.0 \, atm$ then the value of $ln\left(P_{final}\right)$ is
(given $ln \, 2=0.7$ )

• A. $\text{1.75}$
• B. $\text{0.176}$
• C. $\text{1.0395}$
• D. $\text{2.0}$

Answer 1

Answer: (A)

Isoentropic process is an adiabatic process. So,
$\left(\frac{T_{1}}{T_{2}}\right)^{\gamma }=\left(\frac{P_{1}}{P_{2}}\right)^{\gamma - 1}\Rightarrow \left(\frac{300}{600}\right)^{\frac{5}{3}}=\left(\frac{P_{i}}{P_{f}}\right)^{\frac{2}{3}}$
$\Rightarrow \left(\frac{1}{2}\right)^{\frac{5}{3}}=\left(\frac{1}{P_{f}}\right)^{\frac{2}{3}}$
$\Rightarrow \frac{5}{3}ln \, 2=\frac{2}{3}ln \, P_{f}\Rightarrow lnP_{f}=\frac{5}{2}ln \, 2=\text{1.75}$