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 $1.0atm$ then the value of $ln\left(P_{final}\right)$ is
(given $ln2=0.7$ )

• 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,
${\left(\frac{T_1}{T_2}\right)}^{\gamma }={\left(\frac{P_1}{P_2}\right)}^{\gamma -1}\Rightarrow {\left(\frac{300}{600}\right)}^{\frac{5}{3}}={\left(\frac{P_i}{P_f}\right)}^{\frac{2}{3}}$
$\Rightarrow {\left(\frac{1}{2}\right)}^{\frac{5}{3}}={\left(\frac{1}{P_f}\right)}^{\frac{2}{3}}$
$\Rightarrow \frac{5}{3}ln2=\frac{2}{3}ln{P}_f\Rightarrow ln{P}_f=\frac{5}{2}ln2=\text{1.75}$