One mole of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of \(27°C\). If the work done during the process is \(3\, \ kJ\), the final temperature will be equal to \((C_v = 20 JK^{–1})\)

150K

The correct answer is option 1. 150 K.

We know, from the First law of Thermodynamics,

\(\Delta U = q + W\, \ ------(1)\)

We know, for adiabatic condition,

\(q = 0\)

Thus, equation \((1)\) becomes

\(\Delta U = W\, \ -----(2)\)

At constant volume

\(\Delta U = nC_v\Delta T\, \ -----(3)\)

From equation \((1)\) and \((3)\) wehave

\(W = nC_v\Delta T\, \ -------(4)\)

Given,

\(T_i = 27^oC = 27 + 273 = 300\, \  K\)

\(T_f = T\, \ (say)\)

Thus,

\(\Delta T  = T_i - T_f\)

or, \(\Delta T  = 300 - T\)

\(C_v = 20 JK^{–1}\)

\(W = -3\, \ kJ = -3 \times 10^3\) (∵ Work done by the gas is negative)

Substituting these values in equation \((4)\), we get

\(- 3000 = 1 \times 20 \times (T - 300)\)

\(⇒ T - 300 = -150\)

\(⇒ T = 300 - 150\)

\(⇒ T = 150\, \ K\)