Question

What is the work done against the atmosphere when 25 grams of water vaporises at 73 K against a constant external pressure of 1 atm? Assume that stream obeys perfect gas bove laws. Given that the molar enthalpy of vaporisation is $9.72$ Kcal/mole. What is the change of internal energy in the above process?

• A. 1294.0 cal, 11247 cal
• B. 921.4 cal, 11074 cal
• C. -1029.1 cal, 12470 cal
• D. 1129.3 cal, 10207 cal

Answer 1

Answer: (C)

-1029.1 cal, 12470 cal

Volume of $25gm$ of water at $373K$ and $1atm$,

$V_1=0.018L\times \frac{25}{18}$

Volume of $25gm$ of water in vapour state at $373K$,

$V_2=\frac{nRT}{P}=\frac{25\times 0.0821\times 373}{18\times 1}=42.53L$

Now work done, $W=-P(V_2-V_1)=-1(42.53-0.025)Latm$

$=-42.505Latm$

$=-4305.7J$

$=-1029.1cal$

Change in internal energy$=\Delta H_{vap}+W$

$=9.27\times {10}^{3}cal\times \frac{25}{18}-1029.1$

$=12470.9cal$