QJEEM25Phecfd: JEE Main 2025 Physics - Class 11 JEEM25Phecfd Solution: An ideal gas initially at temperature, is compressed suddenly…

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

$\gamma = \frac{c_p}{c_v} = \frac{3}{2}\,.$

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

$T\, V^{\gamma-1} = \text{constant}\,.$

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.

Here are the steps:

Write the adiabatic relation between the initial and final states:

$T_i\, V_i^{\gamma-1} = T_f\, V_f^{\gamma-1}\,.$

Since the gas is compressed to one-fourth of its volume, we have

$V_f = \frac{1}{4} V_i\,.$

Substitute this into the adiabatic relation:

$273\,V_i^{\gamma-1} = T_f \left(\frac{1}{4} V_i\right)^{\gamma-1}\,.$

With $\gamma = \frac{3}{2},$

$\gamma - 1 = \frac{3}{2} - 1 = \frac{1}{2}\,.$

So the relation becomes:

$273\,V_i^{1/2} = T_f \left(\frac{1}{4}\right)^{1/2} V_i^{1/2}\,.$

Cancel the common factor $V_i^{1/2}$ (provided it is nonzero) and simplify:

$273 = T_f \left(\frac{1}{2}\right)\,.$

Now solve for the final temperature $T_f$ :

$T_f = 273 \times 2 = 546\,\mathrm{K}\,.$

The change in temperature is then:

$\Delta T = T_f - T_i = 546 - 273 = 273\,\mathrm{K}\,.$

So, the change in temperature due to the thermodynamic process is $\boxed{273\,\mathrm{K}}.$