Chemical Thermodynamics Question 410
Question: Temperature of 5 moles of a gas is decreased by 2K at constant pressure. Indicate the correct statement
Options:
A) Work done by gas is = 5 R
B) Work done over the gas is = 10 R
C) Work done by the gas = 10 R
D) Work done =0
Show Answer
Answer:
Correct Answer: B
Solution:
- For 5 moles of gas at temperature T, $ PV_1=5RT $ For 5 moles of gas at temperature $ T-2 $ , $ PV_2=5R( T-2 ) $
$ \therefore P(V_2-V_1)=5R( T-2-T ); $
$ P\Delta V=-10R, $
$ -P\Delta V=10R $ When $ \Delta V $ is negative, W is + ve.
