Chemical Thermodynamics Question 338
One mole of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of $ 27^{o}C $ . If the work done during the process is 3 kJ, the final temperature will be equal to $ (C_{v}=20,J{{K}^{-1}}mol^{-1}}$
Options:
A) 150 K
B) 100 K
C) $ {{26.85}^{o}}C $
D) 295 K
Show Answer
Answer:
Correct Answer: A
Solution:
- We know that work done, $ W=C_{v}(T_2-T_1) $
$ 3\times 1000=20,(300-T_2) $ ;
$ \therefore 3000=6000-20,T_2 $
$ \therefore T_2=\frac{3000}{20}=150,K $ .
BETA
