Question

One mole of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of $27^{\circ} C$ The work done is $3 \,kJ$. The final temperature of the gas is equal to $\left[C_{v}=20\, K J^{-1}\right]:$

• A. 75 K
• B. 150 K
• C. 225 K
• D. 300 K

Answer 1

Answer: (B)

150 K

Answer 2

From first law of thermodynamics, $\Delta U = q + W$ As we know that, for adiabatic condition, $q =0$ $\therefore \Delta U = W \ldots . .(1)$ At constant volume, $\Delta U = n C _{ v } \Delta T \ldots . .(2)$ From eq $^{ n }(1) \&(2)$, we have $W = n C _{ v } \Delta T \ldots . .(3)$ Given:- $T _{ i }=27^{\circ} C =(27+273) K =300 K$ $T _{ f }= T ($ say $)=$ ? $\Delta T = T _{ f }- T _{ i }=( T -300)$ $C _{ v }=20 J / K -$ mol $W =-3 kJ =-3 \times 10^{3} J [\because$ Work done by the gas is negative $]$ $T -3000=1 \times 20 \times( T -300)$ $\Rightarrow T =300-150=150 K$