Thermodynamics Question 314
Question: A diatomic ideal gas is heated at constant volume until the pressure is doubled and again heated at constant pressure until volume is doubled. The average molar heat capacity for whole process is:
Options:
A) $ \frac{13R}{6} $
B) $ \frac{19R}{6} $
C) $ \frac{23R}{6} $
D) $ \frac{17R}{6} $
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Answer:
Correct Answer: B
Solution:
[b]Let initial pressure, volume, temperature be $ P _{0},V _{0},T _{0} $ indicated by state A in P-V diagram.
The gas is then isochoric ally taken to state $ B(2P _{0},V _{0},2T _{0}) $
and then taken from state B to state $ C(2P _{0},2V _{0},4T _{0}) $ isobaric ally.
Total heat absorbed by 1 mole of gas
$ \Delta Q=C _{v}(2T _{0}-T _{0})+C _{P}(4T _{0}-2T _{0})=\frac{5}{2} $
$ RT _{0}+\frac{7}{2}R\times 2T _{0}=\frac{19}{2}RT _{0} $
Total change in temperature from state A to C is $ \Delta T=3T _{0} $
$ \therefore $ Molar heat capacity
$ =\frac{\Delta Q}{\Delta T}=\frac{\frac{19}{2}RT _{0}}{3T _{0}}=\frac{19}{6}R. $
BETA
