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?
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Answer
Data: $T _{ f }=2 T _{ i }$, monatomic gas $\therefore \gamma=5 / 3$ $P_{ i } V_{ i }^\gamma=P_{ f } V_{ f }^\gamma$ in an adiabatic process
Now, $P V=n R T \quad \therefore V=\frac{n R T}{P}$
$\therefore V_{ i }=\frac{n R T_{ i }}{P_{ i }}$ and $V_{ f }=\frac{n R T_{ f }}{P_{ f }}$
$\therefore P_{ i }\left(\frac{n R T_{ i }}{P_{ i }}\right)^\gamma=P_{ f }\left(\frac{n R T_{ f }}{P_{ f }}\right)^\gamma$
$ \therefore P_{ i }^{1-\gamma} T_{ i }^\gamma=P_{ f }^{1-\gamma} T_{ f }^\gamma \quad \therefore\left(\frac{T_{ f }}{T_{ i }}\right)^\gamma=\left(\frac{P_{ i }}{P_{ f }}\right)^{1-\gamma}$
$\therefore\left(\frac{T_{ f }}{T_{ i }}\right)^\gamma=\left(\frac{P_{ f }}{P_{ i }}\right)^{\gamma-1} \quad \therefore 2^{5 / 3}=\left(\frac{P_{ f }}{P_{ i }}\right)^{5 / 3-1}=\left(\frac{P_{ f }}{P_{ i }}\right)^{2 / 3}$
$\therefore \frac{5}{3} \log 2=\frac{2}{3} \log \frac{P_{ f }}{P_{ i }} \quad \therefore \frac{5}{3} \times 0.3010=\frac{2}{3} \log \left(\frac{P_{ f }}{P_{ i }}\right)$
$\therefore(2.5)(0.3010)=\log \left(\frac{P_{ f }}{P_{ i }}\right) \therefore 0.7525=\log \left(\frac{P_{ f }}{P_{ i }}\right)$
$\therefore \frac{P_{ f }}{P_{ i }}=\operatorname{antilog} 0.7525=5.656 $
This is the ratio of the final pressure $\left(P_f\right)$ to the initial pressure $\left(P_i\right)$.
Now, $P V=n R T \quad \therefore V=\frac{n R T}{P}$
$\therefore V_{ i }=\frac{n R T_{ i }}{P_{ i }}$ and $V_{ f }=\frac{n R T_{ f }}{P_{ f }}$
$\therefore P_{ i }\left(\frac{n R T_{ i }}{P_{ i }}\right)^\gamma=P_{ f }\left(\frac{n R T_{ f }}{P_{ f }}\right)^\gamma$
$ \therefore P_{ i }^{1-\gamma} T_{ i }^\gamma=P_{ f }^{1-\gamma} T_{ f }^\gamma \quad \therefore\left(\frac{T_{ f }}{T_{ i }}\right)^\gamma=\left(\frac{P_{ i }}{P_{ f }}\right)^{1-\gamma}$
$\therefore\left(\frac{T_{ f }}{T_{ i }}\right)^\gamma=\left(\frac{P_{ f }}{P_{ i }}\right)^{\gamma-1} \quad \therefore 2^{5 / 3}=\left(\frac{P_{ f }}{P_{ i }}\right)^{5 / 3-1}=\left(\frac{P_{ f }}{P_{ i }}\right)^{2 / 3}$
$\therefore \frac{5}{3} \log 2=\frac{2}{3} \log \frac{P_{ f }}{P_{ i }} \quad \therefore \frac{5}{3} \times 0.3010=\frac{2}{3} \log \left(\frac{P_{ f }}{P_{ i }}\right)$
$\therefore(2.5)(0.3010)=\log \left(\frac{P_{ f }}{P_{ i }}\right) \therefore 0.7525=\log \left(\frac{P_{ f }}{P_{ i }}\right)$
$\therefore \frac{P_{ f }}{P_{ i }}=\operatorname{antilog} 0.7525=5.656 $
This is the ratio of the final pressure $\left(P_f\right)$ to the initial pressure $\left(P_i\right)$.
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