You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Heat of Reaction

The molar ent...

Question

The molar enthalpy change for $H_{2} O (l) ⇌ H_{2} O (g)$ at $373$ K and $1$ atm is $41 k J / m o l$ . Assuming ideal behavior, the internal energy change for vaporization of $1$ mol of water at $373 K$ and $1$ atm in kJ mol $^{- 1}$ is:

30.2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

41.0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

48.1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

37.9

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $37.9$
$Δ H = Δ U + Δ (P V)$

$Δ H = Δ U + R T Δ n$

$41 = Δ U + 8.314 \times 373 \times 1 \times 10^{- 3}$

$Δ U = 37.9 k J / m o l$

Hence option $D$ is correct.

Suggest Corrections

Similar questions

Q. The molar enthalpy change for

H_{2} O (1) ⇌ H 2 O (g)

373

K and

1

atm is

41

kJ/mol. Assuming ideal behavior, the internal energy change for vaporization of

1

mol of water at

373

K and

1

atm in kJ

m o l^{- 1}

?

Q. Assuming that water vapour is an ideal gas, the internal energy change

(Δ U)

when

1

mol of water is vaporized at

1

bar pressure and

100^{o}

C (given: molar enthalpy of vaporization of water

41

k J

m o l^{- 1}

1

bar and

373

K and

R = 8.3

m o l^{- 1}

m o l^{- 1}

) will be:

Q. Assuming that, water vapour is an ideal gas, the internal energy change

(Δ U

) when

1

mol of water is vaporised at

1

bar pressure and

100^{o} C

, (given: molar enthalpy of vaporisation of water at

1

bar and

373 K = 41 k J {m o l}^{- 1}

and

R = 8.314 J K^{- 1} {m o l}^{- 1}

) will be:

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}

]

Assuming that water vapour is an ideal gas, the internal energy change (∆U) when 1 mol of water is vapourised at 1 bar pressure and 100 $^{\circ}$ C, (given : molar enthalpy of vapourisation of water at 1 bar and 373 K = 41 kJ mol $^{- 1}$ and R = 8.3 J mol $^{- 1}$ K $^{- 1}$ ) will be

Join BYJU'S Learning Program

The molar enthalpy change for H2O(l)⇌H2O(g) at 373K and 1 atm is 41 kJ/mol. Assuming ideal behavior, the internal energy change for vaporization of 1 mol of water at 373 K and 1 atm in kJ mol−1 is:

The correct option is D 37.9ΔH=ΔU+Δ(PV)ΔH=ΔU+RTΔn41=ΔU+8.314×373×1×10−3ΔU=37.9 kJ/molHence option D is correct.

The molar enthalpy change for $H_{2} O (l) ⇌ H_{2} O (g)$ at $373$ K and $1$ atm is $41 k J / m o l$ . Assuming ideal behavior, the internal energy change for vaporization of $1$ mol of water at $373 K$ and $1$ atm in kJ mol $^{- 1}$ is:

The correct option is D $37.9$
$Δ H = Δ U + Δ (P V)$

$Δ H = Δ U + R T Δ n$

$41 = Δ U + 8.314 \times 373 \times 1 \times 10^{- 3}$

$Δ U = 37.9 k J / m o l$

Hence option $D$ is correct.