Question

An ideal monatomic gas is adiabatically compressed so that its final temperature is twice its initial temperature. What is the ratio of the final pressure to its initial pressure?

Answer 1

Given : T_f = 2T_i, monatomic gas  ϒ = 5/3

P_iV_i^ϒ = P_fV_f^ϒ in an adiabatic process

Now, PV= nRT  ∴ V = nRT/P

∴ V_i = $\frac{nR{T}_i}{P_i}$ and V_f = $\frac{nR{T}_f}{P_f}$

∴ $P_i(\frac{nR{T}_i}{P_i}{)}^{\gamma }=P_f(\frac{nR{T}_f}{P_f}{)}^{\gamma }$

∴ P_i^1−ϒT_i^ϒ = P_f^1−ϒT_f^ϒ

∴ $(\frac{T_f}{T_i}{)}^{\gamma }$ = $(\frac{P_i}{P_f}{)}^{1-\gamma }$

∴ $(\frac{T_f}{T_i}{)}^{\gamma }$ = $(\frac{P_f}{P_i}{)}^{\gamma -1}$

∴ $2^{\frac{5}{3}}=(\frac{P_f}{P_i}{)}^{\frac{5}{3}-1}=(\frac{P_f}{P_i}{)}^{\frac{2}{3}}$

∴ $\frac{5}{3}log2=\frac{2}{3}log(\frac{P_f}{p_i})$

∴ $\frac{5}{3}0.3010=\frac{2}{3}log(\frac{P_f}{p_i})$

∴ $(2.5)(0.3010)=log(\frac{P_f}{p_i})$

∴ 0.7525 = $log(\frac{P_f}{p_i})$

= $(\frac{P_f}{p_i})$ = antilog (0.7525) = 5.656

This is ratio of final pressure to its initial pressure