Thermodynamics Question 271
Question: The variation of internal energy (U) with the pressure (P) of 2.0 mole gas in cyclic process abcda. The temperature of gas at c and d are 300 K and 500 K. calculate the heat absorbed by the gas during the process.
Options:
A) 400 R In 2
B) 200 R In 2
C) 100 R In 2
D) 300 R In 2
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Change in internal energy for cyclic process $ ( \Delta U )=0. $
For process $ a\to b $ , (P-constant) $ {W _{a\to b}}=P\Delta V $
$ =nR\Delta T=-400R $
For process $ b\to c $ (T-constant) $ {W _{b\to c}}=-2R(300)\text{ln 2} $
For process $ c\to d $ , (P-constant) $ {W _{c\to d}}=+400R $
For process $ d\to a $ , (T-constant) $ {W _{d\to a}}=+2R(500)\text{ ln 2 } $
Net work (AW) $ ={W _{a\to b}}+{W _{b\to c}}+{W _{c\to d}}+{W _{d\to a}} $
$ \Delta W=400R\text{ ln 2} $
$ \therefore dQ=dU+dW $ , first law of thermodynamics $ \therefore dQ=400R\text{ ln }2. $
