Find out the internal energy change for the reaction A l - Ag at 373 K- Heat of vaporisation is 40-66 kJ-mol and R- 8-3 J mol-1 K-1-

Given : A (l) → A(g) nbsp; so, Δ n_g = n_p n_r=1 0=1 Δ H =Δ U + nbsp;Δ n_gRT Δ U =Δ H Δ n_gRT =40660 −1×8.3×373=40660−3095.9=37564 J/mol Δ U=37.56 kJmol ...

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Standard XII

Chemistry

Difference between Internal Energy and Enthalpy

Find out the ...

Question

Find out the internal energy change for the reaction $A (l) \to A (g)$ at $373 K$ . Heat of vaporisation is $40.66 k J / m o l$ and $R = 8.3 J m o l^{- 1} K^{- 1}$ .

50.56 k J m o l^{- 1}

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22.56 k J m o l^{- 1}

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37.56 k J m o l^{- 1}

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45.56 k J m o l^{- 1}

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Solution

The correct option is C $37.56 k J m o l^{- 1}$
Given :
$A (l) \to A (g) s o, Δ n_{g} = n_{p} - n_{r} = 1 - 0 = 1$
$Δ H = Δ U + Δ n_{g} R T$
$Δ U = Δ H - Δ n_{g} R T$
$= 40660 - 1 \times 8.3 \times 373$ $= 40660 - 3095.9 = 37564 J / m o l$
$Δ U = 37.56 k J m o l^{- 1}$

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Similar questions

Q. Find out the internal energy change for the reaction

A (l) \to A (g)

373 K

. Heat of vaporisation is

40.66 k J / m o l

and

R = 8.3 J m o l^{- 1} K^{- 1}

Q. A swimmer coming out from a pool is covered with a film of water weighing about 18 g. Calculate the internal energy of vaporisation at

100^{0} C

[ ∆ vap H for water at 373 K = 40.66 kJ mol -1 ]

Assuming that, water vapour is an ideal gas the internal energy change (ΔU)(ΔU) when 11 mol of water is vaporized at 11 bar pressure and 100o100o C will be:

[given: molar enthalpy of vaporisation of water at

1

bar and

373

= 41

m o l^{- 1}

and

R = 8.3

K^{- 1}

m o l^{- 1}

]

Q. A swimmer coming out from a pool is covered with a film of water weighing about

18

g. How much heat must be supplied to evaporate this water at

298

K? Calculate the internal energy of vaporisation at

100^{o}

Δ_{v a p} H^{⊝}

for water at

373

= 40.66

m o l^{- 1}

Q. For water

Δ_{v a p} H = 41 k J {mol}^{- 1}

373 K

and 1 bar pressure. Assuming that water vapour is an ideal gas the that occupies a much larger volume than liquid water, the internal energy change during evaporation of water is ___

k J {mol}^{- 1}

.
[Use :

R = 8.3 J {mol}^{- 1} K^{- 1}

]

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Difference between Internal Energy and Enthalpy

Standard XII Chemistry

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Find out the internal energy change for the reaction A(l)→A(g) at 373 K. Heat of vaporisation is 40.66 kJ/mol and R=8.3 Jmol−1K−1.

The correct option is C 37.56 kJmol−1Given : A(l)→A(g)so, Δng=np−nr=1−0=1 ΔH=ΔU+ΔngRT ΔU=ΔH−ΔngRT =40660 −1×8.3×373=40660−3095.9=37564 J/mol ΔU=37.56 kJmol−1

Find out the internal energy change for the reaction $A (l) \to A (g)$ at $373 K$ . Heat of vaporisation is $40.66 k J / m o l$ and $R = 8.3 J m o l^{- 1} K^{- 1}$ .

The correct option is C $37.56 k J m o l^{- 1}$
Given :
$A (l) \to A (g) s o, Δ n_{g} = n_{p} - n_{r} = 1 - 0 = 1$
$Δ H = Δ U + Δ n_{g} R T$
$Δ U = Δ H - Δ n_{g} R T$
$= 40660 - 1 \times 8.3 \times 373$ $= 40660 - 3095.9 = 37564 J / m o l$
$Δ U = 37.56 k J m o l^{- 1}$