**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?**

**Given** : T_{f} = 2T_{i}, monatomic gas ϒ = 5/3

P_{i}V_{i}^{ϒ} = P_{f}V_{f}^{ϒ} in an adiabatic process

Now, PV= nRT ∴ V = nRT/P

∴ V_{i }= \(\frac{nRT_i}{P_i}\) and V_{f }= \(\frac{nRT_f}{P_f}\)

∴ \(P_i(\frac{nRT_i}{P_i})^\gamma = P_f(\frac{nRT_f}{P_f})^\gamma\)

∴ P_{i}^{1−ϒ}T_{i}^{ϒ} = P_{f}^{1−ϒ}T_{f}^{ϒ}

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

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

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

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

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

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

∴ 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**