One mole of an ideal gas at an initial temperature of T K does 6R joules of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is 5/3, then final temperature of the gas will be :

Question

One mole of an ideal gas at an initial temperature of T K does 6R joules of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is 5/3, then final temperature of the gas will be : (A) (T - 4) K (B) (T + 4) K (C) (T - 2.4) K (D) (T + 2.4) K

Tardigrade Admin · Accepted Answer

Work done by gas in adiabatic process W=(nR(Ti-Tf)/&gamma; -1) ∴ 6R=(1× R(T-T2)/(5/3)-1) or 6=(T-T2/2/3) or T-T2=4 T2=(T-4)k

Q. One mole of an ideal gas at an initial temperature of T K does 6R joules of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is 5/3, then final temperature of the gas will be :

(T - 4) K

(T + 4) K

(T - 2.4) K

(T + 2.4) K

Work done by gas in adiabatic process $W = \frac{n R ( T _{i} - T _{f} )}{γ - 1}$ $∴$ $6 R = \frac{1 \times R ( T - T _{2} )}{( 5/3 ) ^{- 1}}$ or $6 = \frac{T - T _{2}}{2/3}$ or $T - T_{2} = 4$ $T_{2} = (T - 4) k$

Q. One mole of an ideal gas at an initial temperature of T K does 6R joules of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is 5/3, then final temperature of the gas will be :

(T - 4) K

(T + 4) K

(T - 2.4) K

(T + 2.4) K

Work done by gas in adiabatic process W=γ−1nR(Ti​−Tf​)​ ∴ 6R=(5/3)−11×R(T−T2​)​ or 6=2/3T−T2​​ or T−T2​=4 T2​=(T−4)k

Work done by gas in adiabatic process $W = \frac{n R ( T _{i} - T _{f} )}{γ - 1}$ $∴$ $6 R = \frac{1 \times R ( T - T _{2} )}{( 5/3 ) ^{- 1}}$ or $6 = \frac{T - T _{2}}{2/3}$ or $T - T_{2} = 4$ $T_{2} = (T - 4) k$