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Two moles of an ideal gas undergoes isothermal reversible expansion from 2 L to 8 L at 300 K. The enthalpy change of the gas is: 

(IIT JEE 2004)

a) 4.8 kJ 

b) 11.4 kJ 

c) zero 

d) -11.4 kJ

Logic and Explanation:

We know that, for ideal gases, the mathematical relation of enthalpy, H with internal energy, E and PV can be written as:

H = U + PV

Therefore, change in enthalpy:

ΔH = ΔU + Δ(PV) = ΔU + Δ(nRT)

In the given process, Δn = 0, since there is no change in the number of moles and ΔT = 0, since the process is isothermal.



However, for isothermal process, the change in internal energy, ΔU = 0.


ΔH = ΔU = 0

Note: Since U and PV are constant for isothermal process, H must be constant i.e., ΔH = 0. The pressure is not constant in the above mentioned process. It decreases with increase in the volume at constant temperature (Boyle's law). The pressure must be changed. Otherwise, volume cannot be changed i.e. no pressure-volume work is done if both temperature and pressure are kept constant.

However, ΔH (as well as ΔU) may not be zero for isothermal processes that are involving real gases that are non-ideal.


The correct answer is 'c'

Related questions

2) One mole of a non ideal gas undergoes a change of state(2.0 atm, 3.0 L, 95 K) to (4.0 atm, 5.0 L, 245 K) with a change in internal energy, ΔU = 30 L atm. The change in enthalpy (ΔH) of the process in L atm is:  

(IIT JEE 2002) 

1) 40.0 

2) 42.3

3) 44.0

4) not defined, because pressure is not constant.

Logic & solution: 

ΔH = ΔU + Δ(PV) =  ΔU + (P2V2 - P1V1) = 30.0 L atm + (4.0 atm x 5.0 L) - (2.0 atm x 3.0 L) = 44.0 L atm.


Correct option: 3


1) What is enthalpy?

2) What is internal energy?

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Author: Aditya vardhan Vutturi