The standard molar enthalpy of vaporisation of benzene Δ text vap H° at 353 K is 30.8 kJ mol -1. If the benzene vapours behave as an ideal gas, the change in internal energy of vaporisation of 78 g of benzene at 353 K in kJ mol -1 is

Question

The standard molar enthalpy of vaporisation of benzene Δ text vap H° at 353 K is 30.8 kJ mol -1. If the benzene vapours behave as an ideal gas, the change in internal energy of vaporisation of 78 g of benzene at 353 K in kJ mol -1 is (A) 37.87 (B) 27.87 (C) 33.74 (D) 17.87

Tardigrade Admin · Accepted Answer

Given,

Standard molar enthalpy of benzene

(Δ H) = 30.8 k J m o l^{- 1}

Temperature

(T) = 353 K

Equation

(Liquid) C_{6} H_{6} Δ (Gas) C_{6} H_{6}

Here, change in gaseous mole

(Δ n_{g}) = 1

Now,

Δ H = Δ E + Δ n_{g} RT

30.8 k J m o l^{- 1} = Δ E + (1) (0.082) \times (101.32) (353) 1 0^{- 3} k J m o l^{- 1}

30.8 k J m o l^{- 1} = Δ E + 2934.84 \times 1 0^{- 3} k J m o l^{- 1}

Δ E = 30.8 - 2.93

Δ E = 27.87 k J m o l^{- 1}

Q. The standard molar enthalpy of vaporisation of benzene Δvap ​H∘ at 353K is 30.8kJmol−1. If the benzene vapours behave as an ideal gas, the change in internal energy of vaporisation of 78g of benzene at 353K in kJmol−1 is

37.87

27.87

33.74

17.87

Q. The standard molar enthalpy of vaporisation of benzene $Δ_{vap} H^{\circ}$ at $353 K$ is $30.8 k J m o l^{- 1}$ . If the benzene vapours behave as an ideal gas, the change in internal energy of vaporisation of $78 g$ of benzene at $353 K$ in $k J m o l^{- 1}$ is