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One mole of an ideal diatomic gas undergoes a process described by the equation $\mathrm{P V^{\frac{2}{3}}=}$ constant. The initial temperature and volume of the gas are $\mathrm{T_1}$ and $\mathrm{V_1}$ , respectively. The gas is taken through two steps: in Step 1, it undergoes an isobaric process, and in Step 2, it undergoes an isochoric process. The final temperature of the gas is $\mathrm{T_2}$ . Which of the following is true?
Option: 1
$\mathrm{T_2=2 T_1}$

Option: 2
$\mathrm{T_2=\frac{2}{3} T_1}$

Option: 3
$\mathrm{T_2=\frac{3}{2} T_1}$

Option: 4
$\mathrm{T_2=4 T_1}$

Answers (1)

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For the first step, the process is described by the equation $\mathrm{P V^{\frac{2}{3}}=}$ constant. Since this is an isobaric process ( P is constant ), we can write:

$\mathrm{ \begin{gathered} V_1^{\frac{2}{3}}=V_2^{\frac{2}{3}} \\ V_1=V_2 \end{gathered} }$
For the second step, the process is isochoric (constant volume), so $\mathrm{V_2=V_1}$ . Now, let's find the final temperature $\mathrm{\left(T_2\right)}$ of the gas after the two steps. We have:

$\mathrm{ \frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2} }$

Since $\mathrm{V_2=V_1}$ and $\mathrm{P_1=P_2}$ (isobaric process), we get:

$\mathrm{ \frac{V_1}{T_1}=\frac{V_1}{T_2} }$

$\mathrm{ T_2=T_1 }$

So, the correct answer is: 3) $\mathrm{T_2=\frac{3}{2} T_1}$

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