The internal energy of an ideal gas does not change in a reversible __________ process.
A. Isothermal
B. Adiabatic
C. Isobaric
D. Isometric
Answer: Option A
Solution(By Examveda Team)
We know for an ideal gas the internal energy is given by:Actually the internal energy (U) of any substance is a function of
$$\eqalign{ & dU = CvdT - \left[ {P + T\left( {\frac{{\left( {\frac{{\partial V}}{{\partial T}}} \right)p}}{{\left( {\frac{{\partial V}}{{\partial P}}} \right)T}}} \right)dV} \right] \cr & {\text{For an ideal gas, }}PV = RT \cr & {\text{So, }}\left( {\frac{{\partial V}}{{\partial T}}} \right)p = \frac{R}{P}\,\,{\text{and }}\left( {\frac{{\partial V}}{{\partial P}}} \right)T = - \frac{{RT}}{{{P^2}}} \cr & {\text{Hence, }}dU = CvdT \cr} $$
So, in case of isothermal process the internal energy remains constant.
Related Questions on Chemical Engineering Thermodynamics
A. Maxwell's equation
B. Thermodynamic equation of state
C. Equation of state
D. Redlich-Kwong equation of state
Henry's law is closely obeyed by a gas, when its __________ is extremely high.
A. Pressure
B. Solubility
C. Temperature
D. None of these
A. Enthalpy
B. Volume
C. Both A & B
D. Neither A nor B
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