An ideal gas initially at $0^{\circ} \mathrm{C}$ temperature, is compressed suddenly to one fourth of its volume. If the ratio of specific heat at constant pressure to that at constant volume is $3 / 2$, the change in temperature due to the thermodynamic process is _________ K.

<p>We’re given an ideal gas initially at $T_i = 0^\circ \mathrm{C} = 273\,\mathrm{K}.$ The gas is suddenly compressed to one-fourth of its initial volume, and we are told that</p> <p>$\gamma = \frac{c_p}{c_v} = \frac{3}{2}\,.$</p> <p>For an adiabatic process (one in which there is no heat exchange), if the process were reversible, the relation between the temperature and volume would be</p> <p>$T\, V^{\gamma-1} = \text{constant}\,.$</p> <p>Even though the compression is “sudden” (and hence irreversible), the final equilibrium state of the gas is uniquely determined by its internal energy (which is a state function). That allows us to use the adiabatic relation between the initial and final states.</p> <p>Here are the steps:</p> <p><p>Write the adiabatic relation between the initial and final states:</p> <p>$T_i\, V_i^{\gamma-1} = T_f\, V_f^{\gamma-1}\,.$</p></p> <p><p>Since the gas is compressed to one-fourth of its volume, we have</p> <p>$V_f = \frac{1}{4} V_i\,.$</p></p> <p><p>Substitute this into the adiabatic relation:</p> <p>$273\,V_i^{\gamma-1} = T_f \left(\frac{1}{4} V_i\right)^{\gamma-1}\,.$</p></p> <p><p>With $\gamma = \frac{3}{2},$</p> <p>$\gamma - 1 = \frac{3}{2} - 1 = \frac{1}{2}\,.$</p> <p>So the relation becomes:</p> <p>$273\,V_i^{1/2} = T_f \left(\frac{1}{4}\right)^{1/2} V_i^{1/2}\,.$</p></p> <p><p>Cancel the common factor $V_i^{1/2}$ (provided it is nonzero) and simplify:</p> <p>$273 = T_f \left(\frac{1}{2}\right)\,.$</p></p> <p><p>Now solve for the final temperature $T_f$:</p> <p>$T_f = 273 \times 2 = 546\,\mathrm{K}\,.$</p></p> <p><p>The change in temperature is then:</p> <p>$\Delta T = T_f - T_i = 546 - 273 = 273\,\mathrm{K}\,.$</p></p> <p>So, the change in temperature due to the thermodynamic process is $\boxed{273\,\mathrm{K}}.$</p>