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Chemistry
1 mole of NH 3 gas at 27° C is expanded in reversible adiabatic condition to make volume 8 times (γ=1.33). Final temperature and work done respectively are :
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Q. $1$ mole of $NH _{3}$ gas at $27^{\circ} C$ is expanded in reversible adiabatic condition to make volume $8$ times $(\gamma=1.33)$. Final temperature and work done respectively are :
A
150 K , 900 cal
80%
B
150 K , 400 cal
4%
C
250 K , 1000 cal
16%
D
200 K , 800 cal
0%
Solution:
$T_{1} V_{1}^{\gamma-1}=T_{2} V_{2}^{\gamma-1} \Rightarrow 300 \times V^{1 / 3}$
$=T_{2}(8 V)^{1 / 3} \Rightarrow T_{2}=150\, K$
$W=n C_{v}\left(T_{2}-T_{1}\right)=1 \times 3 R(150-300)$
$=3 \times 2(-150)=-900 \,ca$