Question

One mole of a mono-atomic gas behaving as per PV = nRT at 27°C is subjected to reversible isoentropic compression untill final temperature reaches 327°C. If the initial pressure was 1.0 atm then the value of lnP (final) is –

(givenln 2 = 0.7)

• A. 1.75
• B. 0.176
• C. 1.0395
• D. 2.0

Answer 1

Answer: (A)

1.75

Answer 2

Isoentropic process is adiabatic process. So

$\left( \frac{T_{1}}{T_{2}} \right)^{\gamma} = \left( \frac{P_{1}}{P_{2}} \right)^{\gamma –1} \Rightarrow \left( \frac{300}{600} \right)^{5/3} = \left( \frac{P_{i}}{P_{f}} \right)^{2/3}$

̃ $\left( \frac{1}{2} \right)^{5/3} = \left( \frac{1}{P_{f}} \right)^{2/3}$

̃ $\frac{5}{3}$ ln 2 = $\frac{2}{3}$ln P_f ̃ lnP_f = $\frac{5}{2}$ln 2 = 1.75