A reversible isothermal evaporation of 90 g of water is carried out at 100° C. Heat of evaporation of water is 9.72 kcal mol -1. Assuming water vapour to behave like an ideal gas, what is the change in internal energy of the system?

Question

A reversible isothermal evaporation of 90 g of water is carried out at 100° C. Heat of evaporation of water is 9.72 kcal mol -1. Assuming water vapour to behave like an ideal gas, what is the change in internal energy of the system? (A) 48.6 kcal (B) 52.23 kcal (C) 44.87 kcal (D) 56.06 kcal

Tardigrade Admin · Accepted Answer

Change in enthalpy (Δ H)= Heat of evaporation × Number of moles =9.72 × 5 Δ H=Δ E+Δ n R T Δ E=48.6-(5 × 2 × 10-3 × 373) =44.87 kcal

Q. A reversible isothermal evaporation of $90 g$ of water is carried out at $10 0^{\circ} C$ . Heat of evaporation of water is $9.72 k c a l m o l^{- 1}$ . Assuming water vapour to behave like an ideal gas, what is the change in internal energy of the system?

$48.6 k c a l$

$52.23 k c a l$

$44.87 k c a l$

$56.06 k c a l$

Change in enthalpy $(Δ H) =$ Heat of evaporation $\times$ Number of moles $= 9.72 \times 5$
$Δ H = Δ E + Δ n RT$
$Δ E = 48.6 - (5 \times 2 \times 1 0^{- 3} \times 373)$
$= 44.87 k c a l$

Q. A reversible isothermal evaporation of 90g of water is carried out at 100∘C. Heat of evaporation of water is 9.72kcalmol−1. Assuming water vapour to behave like an ideal gas, what is the change in internal energy of the system?

48.6kcal

52.23kcal

44.87kcal

56.06kcal