Two moles of an ideal gas are expanded isothermally and reversibly from 1 litre to 10 litre at 300K . The enthalpy change (in kJ ) for the process is:

Δ H= Δ U+W

From the formula of enthalpy it is clear that, we have to first calculate the value of work done for the expansion of two moles of an ideal gas from 1 liter to 10 liter at 300K

By using following formula:

W= -2.303nRT logVFVI

Where,

n = No. of moles which is equal to 2 according to the given question.

T = Temperature of the reaction which is equal to 300K

Vf = Final volume of gas which is equal to 10 liter.

Vi = Initial volume of gas which is equal to 1 liter according to the question.

R = Universal gas constant whose value is equal to 8.314J/Kmol

Now putting these values on the above equation of the work done and we get, W = - 2.30328.314300log101 W= - 2.30328.3143001 since log101=1

W= - 11488.285J

As we know that enthalpy change for the given reaction is calculated as:

Δ H= Δ U+W

Δ H= Δ U+W

And for an isothermal reaction change in internal energy is always zero i.e. Δ U=0, so the value of change in enthalpy is equal to the value of work done during the expansion.

i.e Δ H=W

H=-11.488kJ

Hence option (B) is correct i.e. change in enthalpy is equal to −11.4