Question

5 moles of a liquid ${}^{\prime }{L}^{\prime }$ are converted into vapour at its boiling point ${273}^{o}C$ and at a pressure of 1 atm. If the value of latent heat of vapourisation of liquid ${}^{\prime }{L}^{\prime }$ is $273Latm/mol$, then which of the following statements is/are correct?
[Assume volume of the liquid to be negligible and vapour of the liquid to behave ideally]

• A. Work done by the system in the above process is $-224Latm$
• B. The enthalpy change $\left(\Delta H\right)$ for the above process is $1365Latm$
• C. The entropy of the system increases by $2.5Latm/K$ in the above process
• D. The value of $\Delta U$ for the above process is $1589Latm$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A Work done by the system in the above process is $-224Latm$
B The enthalpy change $\left(\Delta H\right)$ for the above process is $1365Latm$
C The entropy of the system increases by $2.5Latm/K$ in the above process
D The value of $\Delta U$ for the above process is $1589Latm$

(A) The volume of liquid is very small when compared to the volume of vapour.
Hence, when a liquid vapourizes, the change in the volume can be approximated to the volume of vapour.
Hence, the work done during phase transition will be $-P\Delta V=-PV=-nRT=-5\times 0.0821\times (273+273)=224$ L atm.
Since, the work is done by the system, it has negative sign.

(B) The latent heat of vaporisation is 273 Latm/mol.
The enthalpy change for the above process (involving 5 moles) is $273Latm/mol\times 5=1365Latm$

(C) The increase in the entropy in the above process is $\frac{\Delta H}{T}=\frac{1365}{273+273}=2.5Latm/K$

(D) The change in the internal energy for the above process is $\Delta U=\Delta H-P\Delta V=1356-(-224)=1589Latm$