For water at $100^\circ$C and 1 bar $\Delta_{\text{vap}}H - \Delta_{\text{vap}}U =$ ______ $\times 10^{2}$ J mol$^{-1}$ (Round off to the Nearest integer) [Use: R = 8.31 J mol$^{-1}$ K$^{-1}$] [Assume volume of $\ce{H2O}$(l) is much smaller than volume of $\ce{H2O}$(g). Assume $\ce{H2O}$(g) can be treated as an ideal gas]

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Chemical Thermodynamics

For water at

10 0^{\circ}

C and 1 bar

Δ_{vap} H - Δ_{vap} U =

______

\times 1 0^{2}

J mol

^{- 1}

(Round off to the Nearest integer)
[Use: R = 8.31 J mol

^{- 1}

^{- 1}

]
[Assume volume of

H X_{2} O

(l) is much smaller than volume of

H X_{2} O

(g). Assume

H X_{2} O

(g) can be treated as an ideal gas]

Answer: 31

For vaporisation at constant pressure,

Δ_{vap} H - Δ_{vap} U = P (V_{g} - V_{l})

The volume of the liquid is negligible beside that of the vapour, and the vapour is ideal, so

P V_{g} = n_{g} R T

with one mole of gas formed per mole of water vaporised:

Δ_{vap} H - Δ_{vap} U = R T = 8.31 \times 373 = 3099.6 J mo l^{- 1}

3099.6 J mo l^{- 1} = 30.996 \times 1 0^{2} J mo l^{- 1}

Rounded to the nearest integer, the required value is

31 \times 1 0^{2}

mo l^{- 1}

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Integer answer