For a constant volume process

A. dE = C_pdT

B. dE = C_vdT

C. dQ = dE + pdV

D. dW = pdV

Answer: Option B

Solution(By Examveda Team)

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
$$\eqalign{ & \delta Q = dU \cr & \Rightarrow dU = CvdT. \cr} $$