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During Joule-Thomson expansion of gases
Answer & Solution
Correct Answer:
Option
A
During, joule-thomson effect basically the change in internal (inter-molecular) energy and change in pressure energy will be equal to zero hence, enthalpy will remain constant.
$$\eqalign{ & dU + d\left( {PV} \right) = 0 \cr & \Rightarrow d\left( {U + PV} \right) = 0 \cr & \Rightarrow d\left( H \right) = 0 \cr & {\text{or }}H = {\text{constant}}{\text{.}} \cr} $$
$$\eqalign{ & dU + d\left( {PV} \right) = 0 \cr & \Rightarrow d\left( {U + PV} \right) = 0 \cr & \Rightarrow d\left( H \right) = 0 \cr & {\text{or }}H = {\text{constant}}{\text{.}} \cr} $$
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