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For isothermal reversible expansion of an ideal gas:
A: ΔH>0 and ΔU=0
B: ΔH>0 and ΔU<0
C: ΔH=0 and ΔU=0
D: ΔH=0and ΔU>0

Answer
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Hint: Isothermal expansion is change in the system in which temperature remains constant. In such an expansion change in the system will be very slow that there will be enough time to adjust temperature so that initial temperature will be equal to final temperature.

Complete step by step solution:
H represents enthalpy of a system. It is calculated as
H=U+PV
ΔH represents change in enthalpy of system which is calculated as
ΔH=ΔU+PdV+VdP (taking derivative both sides give change in enthalpy)
U represents internal energy and ΔH represents change in internal energy.
ΔU can be found by equation
ΔU=CvΔT here ΔT is the change in temperature and as given in the question statement it is clear that process is isothermal that is there is no change in temperature throughout the process so change in internal energy ΔU is found to be zero
Since ΔH can be found by the relation ΔH=ΔU+PdV+VdP which can also be written as ΔH=ΔU+RΔT (by ideal gas equation PV=nRT) from above interpretation ΔU comes to be zero and ΔT also comes to be zero as temperature is constant and process is isothermal so change in enthalpy (ΔH) also come to be zero.
So our correct option is C that is ΔH=0 and ΔU=0
Additional information:
Reversible process: It is a process in which the system and surroundings can be returned to the original conditions from the final state.
Irreversible process: It is a process that cannot return both the system and the surroundings to the original conditions from the final state.

Note: Since we cannot find the absolute internal energy or enthalpy of gas so we deal with the change of internal energy of a system through different states. There are some processes carried out to change one state of the system to another. such as isothermal(Temperature constant), isobaric(Pressure constant), isochoric(Volume constant), reversible, irreversible etc.