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dU for Isothermal Process for Vander Waals Gas and Ideal Gas respectively are:
Correct Answer : a
Explanation : a) U = (v,T)
dU = (du/dv)T dv + (dU/dT) dT
here, (dU/dT)_{v }= nC_{v} and (dU/dV)T = an^{2}/v^{2}
for isothermal process dT = 0
dU = an^{2}/v^{2} dv
Integrating it,
∫(U_{2} to U_{1})dU = ∫(V_{2} to V_{1})an^{2}/v^{2} dv
(U_{2}  U_{1}) = an^{2}[1/v](V_{2} to V_{1})
ΔU = an2[1/v^{2}  1/v^{1}]
b)we can express the relationship between internal energy and temperature as:
ΔU = CvdT
internal energy is a function of temperature because internal energy of ideal gas comprises of molecular kinetic energy which further depends on the temperature and hence,
For isothermal process, dT = 0, then ΔU = 0
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