Question

10 moles of a liquid L is $50\%$ converted into its vapor at its boiling point $\left({273}^{o}C\right)$ and at a pressure of 1 atm. If the value of the latent heat of vaporization of liquid L is 273 L atm/mol, then which of the following statements is/are correct?

(Assume volume of liquid to be negligible and vapor of the liquid to behave ideally)

• A. Work done by the system in the above process is 224 L atm.
• B. The enthalpy change $\left(\Delta H\right)$ for the above process is 1365 L atm (with respect to magnitude only).
• C. The entropy of the system increases by 2.5 L atm in the above process.
• D. The value of $\Delta U$ for the above process is 1589 L atm.

Answer 1

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

The correct options are
A Work done by the system in the above process is 224 L atm.
B The enthalpy change $\left(\Delta H\right)$ for the above process is 1365 L atm (with respect to magnitude only).
C The entropy of the system increases by 2.5 L atm in the above process.

$V_1=5\times R\times 546=224L$
$W=-P_{ext}\left(\Delta V\right)=-1\left(224L\right)=-224$ L atm
$\therefore$ Work done by system $=224$ L atm
Enthalpy change $\left(\Delta H\right)=q=273\times 5=1365$ L atm
$\Delta S=\frac{\Delta H_{vap}}{T}=\frac{1365}{546}=2.5$ L atm/K

$\Delta U=q+W=1365-224=1141$ L atm