Chemical Thermodynamics Question 290
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 $ , then final temperature of the gas is $ (C_{V}=20,J/mol\cdot K) $ [Pb. CET 2002]
Options:
A) 100 K
B) 150 K
C) 195 K
D) 255 K
Show Answer
Answer:
Correct Answer: B
Solution:
Given number of moles =1 Initial temperature $ =27^{o}C=300K $ Work done by the system $ =3,KJ=3000J $ It will be $ (-) $ because work is done by the system. Heat capacity at constant volume $ (Cv)=20,J/K $ We know that work done $ W=-nC_{V},(T_2-T_1) $ ; $ -3000=-1\times 20(T_2-300) $
$ 3000=-20T_2+6000 $
$ 20T_2=3000 $ ; $ T_2=\frac{3000}{20}=150K $
BETA
