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I am trying to solve the following question.

A gas is enclosed in a cylindrical can fitted with a piston. The walls of the can and the piston are adiabatic. The initial pressure, volume and temperature of the gas are 100 kPa, 400 $cm^3$ and 300 K respectively.The ratio of the specific heat capacities of the gas is C$_p$ / C$_v$ = 1.5. Find the pressure and temperature of the gas if it is suddenly compressed to 100 $cm^3$.

I tried to solve it by equating dU and δW. Since the gas is suddenly compressed it is an irreversible adiabatic change. δW = −$p_2dV$ and dU = $nC_vdT$. $p_2$ is the final pressure, $n$ is the number of moles present and $C_v$ is the molar specific heat capacity. (I followed the method given in this answer.)

$$nC_v(T_2−T_1) = −p_2(V_2−V_1)$$ From ideal gas equation $p_2 = \frac{nRT_2}{V_2}$. From the ratio given in the question, $C_v$ comes out to be $2R$. $$2nR(T_2−T_1) = −\frac{nRT_2(V_2−V_1)}{V_2}$$ After dividing both sides by $nR$ and substituting values of $V_1$ and $V_2$, the equation becomes
$$2T_2−2T_1=3T_2$$ This equation is wrong. I have copied everything except the value of $p_2$ from the answer linked above so I think I made some mistake in it. Please tell why my method is wrong and how will the question be solved ?

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If you solve your equation algebraically, you get $$T_2=\frac{2T_1}{\left[3-\frac{V_1}{V_2}\right]}$$At $V_1/V_2=2$, $T_2=2T_1$ and at $V_1/V_2=2.5$, $T_2=4T_1$, but at $V_1/V_2=3$, $T_2$ becomes infinite. So it is not possible to compress the gas to a greater compression ratio than 3 without the pressure and temperature becoming infinite (at least not by applying a constant external pressure). In your problem statement, the compression ratio is 4.

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  • $\begingroup$ Am I correct in inferring that the process is irreversible ? (It is not directly mentioned in the question.) $\endgroup$ – ghoul932 Apr 10 at 17:40
  • $\begingroup$ Yes. The key word is suddenly. $\endgroup$ – Chet Miller Apr 10 at 17:50

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