Question 275: Thermodynamics & KTG

Question

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 $$\frac{1}{8}$$ of its original volume while the gas in B is compressed adiabatically to $$\frac{1}{8}$$ of its original volume. The ratio of final pressure of gas in B to that of gas in A is

Accepted Answer

The final pressure of gas in container A after isothermal compression can be found using the equation of state for an ideal gas, $PV = nRT$, where $P$ is the pressure, $V$ is the volume, $n$ is the number of moles, $R$ is the gas constant, and $T$ is the temperature. For an isothermal process, the temperature $T$ is constant, so the equation becomes $P_1V_1 = P_2V_2$.
The final volume is $\frac{1}{8}$ of the initial volume, so $P_2 = P_1 	imes \frac{V_1}{V_2} = P_1 	imes 8 = 8P_1$.
The final pressure of gas in container B after adiabatic compression can be found using the adiabatic equation for an ideal gas, $PV^\gamma = 	ext{constant}$, where $\gamma$ is the ratio of the heat capacities, which is $\frac{5}{3}$ for a monoatomic gas.
Since $V_2 = \frac{V_1}{8}$, we have $P_2 = P_1 	imes \left(\frac{V_1}{V_2}ight) ^ \gamma = P_1 	imes 8^\frac{5}{3} = P_1 	imes 2^5 = 32P_1$.
The ratio of the final pressure of gas in B to that of gas in A is therefore $\frac{32P_1}{8P_1} = 4$.

Answer

$$\frac{1}{8}$$

Answer

8$$^\frac{3}{2}$$

Answer

4

Answer

8

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