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For a constant volume process
Answer & Solution
Correct Answer:
Option
B
Since, under constant volume process the expansion work is zero, and by the definition of $${C_V}$$ we can write $${C_V} = \mathop {\lim }\limits_{\delta T \to 0} {\left( {\frac{{\delta Q}}{{\delta T}}} \right)_V}$$
By first law of thermodynamics
$$\eqalign{ & \delta Q = dU \cr & \Rightarrow dU = CvdT. \cr} $$
By first law of thermodynamics
$$\eqalign{ & \delta Q = dU \cr & \Rightarrow dU = CvdT. \cr} $$
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