c
અચળ દબાણે \(\frac{{{V_1}}}{{{V_2}}}\,\, = \,\frac{{{T_1}}}{{{T_2}}}\,\,\,\,\,\,\therefore \,\,\frac{V}{{2V}}\,\, = \,\,\frac{{300}}{{{T_2}}}\,\,\,\,\,\therefore \,\,{T_2}\,\, = \,\,\,600\,K\,\,\,\,\,\,\)
\(\therefore\) \(\,\Delta T\,\, = \,\,{T_2} - {T_1}\,\, = \,\,600\,\, - \,\,300\,\, = \,\,300\,K\)
હવે \(W = P \Delta V = \mu R \Delta T = (0.1)(2)(300) = 60 cal\)