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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?

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Given : Tf = 2Ti, monatomic gas  ϒ = 5/3

PiViϒ = PfVfϒ in an adiabatic process

Now, PV= nRT  ∴ V = nRT/P

∴ V= \(\frac{nRT_i}{P_i}\) and V= \(\frac{nRT_f}{P_f}\)

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

∴ Pi1−ϒTiϒ = Pf1−ϒTfϒ

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

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

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

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

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

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

∴ 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

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