For an isothermal reversible compression of an ideal gas
A. Only ΔE = 0
B. Only ΔH =0
C. ΔE = ΔH = 0
D. DQ = dE
Answer: Option C
Solution (By Examveda Team)
Actually the internal energy ($$U$$) of a substance is a function of$$\eqalign{ & dU = CvdT - \left[ {P + T\left( {\frac{{\left( {\frac{{\partial V}}{{\partial T}}} \right)p}}{{\left( {\frac{{\partial V}}{{\partial P}}} \right)T}}} \right)dV} \right] \cr & {\text{For an ideal gas, }}PV = RT \cr & {\text{So, }}\left( {\frac{{\partial V}}{{\partial T}}} \right)p = \frac{R}{P}\,\,{\text{and }}\left( {\frac{{\partial V}}{{\partial P}}} \right)T = - \frac{{RT}}{{{P^2}}} \cr & {\text{Hence, }}dU = CvdT \cr} $$
Similarly we can prove $$dH = {C_p}dT$$
So, enthalpy and internal energy are solely functions of temperature in case of ideal gas hence $$\Delta E = \Delta H = 0$$

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