Consider two containers A and B containing monoatomic gases at the same Pressure (P), Volume (V), and Temperature (T). T

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Consider two containers A and B containing monoatomic gases at the same Pressure (P), Volume (V), and Temperature (T). The gas in A is compressed isothermally to 18 of its original volume, while the…

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Given: Two monoatomic gases initially have the same $P$ , $V$ , and $T$ . Gas A is compressed isothermally to $\frac{V}{8}$ , and gas B is compressed adiabatically to $\frac{V}{8}$ .

Find: The ratio of final pressure of gas B to final pressure of gas A.

For the isothermal compression of gas A:

PV = \text{constant}

So,

P_A' \cdot \frac{V}{8} = PV

Hence,

P_A' = 8P

For the adiabatic compression of gas B:

PV^{\gamma} = \text{constant}

For a monoatomic gas,

\gamma = \frac{5}{3}

Therefore,

P_B' \left(\frac{V}{8}\right)^{\frac{5}{3}} = PV^{\frac{5}{3}}

So,

P_B' = P \cdot 8^{\frac{5}{3}}

Now compute the required ratio:

\frac{P_B'}{P_A'} = \frac{P \cdot 8^{\frac{5}{3}}}{8P} = 8^{\frac{2}{3}} = 4

Therefore, the ratio of final pressure of gas B to that of gas A is $4$ . Hence, the correct option is C.

Note: The source options list shows $\frac{1}{8}$ as option C, while the working gives the value $4$ . The solution explicitly marks option C, but by the actual calculation the defensible answer corresponds to the value $4$ .

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