Adiabatic Process

An adiabatic process exchanges no heat, and for an ideal gas it obeys $T{V}^{\gamma -1}=\text{constant}$, so $T_f=T_i{\left(\frac{V_i}{V_f}\right)}^{\gamma -1}$. Here the gas is compressed, so $V_i/V_f=4$ and $\gamma -1=0.4$. Then $T_f=300\times 4^{0.4}$. Evaluating $4^{0.4}=e^{0.4\ln 4}=e^{0.4(1.386)}=e^{0.554}\approx 1.74$, giving $T_f\approx 300\times 1.74=522\text{ K}$. The value 418 K uses an exponent near 0.25, mismatching $\gamma$. The value 600 K assumes temperature scales linearly with volume ratio, ignoring the exponent. The value 300 K wrongly treats the process as isothermal, but no heat leaves the insulated cylinder so the temperature must change. The relation $T{V}^{\gamma -1}=\text{constant}$ follows from combining the adiabatic law $P{V}^{\gamma }=\text{constant}$ with the ideal-gas equation $PV=nRT$ to eliminate pressure. An equivalent route uses the First Law with $Q=0$, so the work done on the gas equals its rise in internal energy $n{C}_V\Delta T$, which again forces the temperature upward. As a physical check, adiabatic compression always heats a gas because work done on it raises its internal energy, so the final temperature exceeding 300 K is exactly what we expect.