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Standard XII

Chemistry

Relation between Cp and Cv

A sample of ...

Question

A sample of $3.0$ moles of perfect gas at $200$ K and $2.0$ atm is compressed reversibly and adiabatically until the temperature reaches $250$ K. Given that molar heat capacity is $27.5$ J K $^{- 1}$ mol $^{- 1}$ at constant volume. Calculate $q, w,$ $Δ U$ , $Δ H$ and the final pressure and volume.

P = 6.28

atm,

V = 11.75

litre,

w = 5.158

kJ,

q =

Δ U = 5.157

kJ,

Δ H = 5.672

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P = 5.28

atm,

V = 11.85

litre,

w = 4.158

kJ,

q =

Δ U = 5.157

kJ,

Δ H = 5.672

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P = 5.26

atm,

V = 11.81

litre,

w = 4.157

kJ,

q =

Δ U = 4.157

kJ,

Δ H = 5.372

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P = 5.26

atm,

V = 11.82

litre,

w = 4.152

kJ,

q =

Δ U = 4.157

kJ,

Δ H = 5.372

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Solution

The correct option is D $P = 5.26$ atm, $V = 11.81$ litre, $w = 4.157$ kJ, $q =$ 0, $Δ U = 4.157$ kJ, $Δ H = 5.372$ kJ
The initial values are $n = 3 m o l e, T_{1} = 200 K, P_{1} = 2.0 a t m, C_{v} = 27.5 J K^{- 1} m o l^{- 1}$
The final values (after compression) are $T_{2} = 250 K, P_{2} = ?$

The heat capacity at constant pressure is $C_{p} = 27.5 + 8.314 J K^{- 1} m o l^{- 1} = 35.814 J K^{- 1} m o l^{- 1}$

Also $γ = \frac{C_{p}}{C_{v}} = \frac{35. 814}{27.5} = 1.30$

But $p_{1}^{1 - γ} \cdot T_{1}^{γ} = P_{2}^{1 - γ} \cdot T_{2}^{γ}$

or $P_{2}^{1 - γ} = P_{1}^{1 - γ} {(\frac{T_{1}}{T_{2}})}^{γ}$

Substitute values in the above expression.
$(P_{2})^{- 0.3} = (2)^{- 0.3} \times {(\frac{200}{250})}^{1.30}$

Thus, $P_{2} = 5.26 a t m$
The initial volume is $V_{1} = \frac{n R T}{P_{1}} = \frac{3 \times 0.0821 \times 200}{2} = 24.63 l i t r e$

Also, $P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ}$

or ${[\frac{V_{2}}{V_{1}}]}^{γ} = \frac{P_{1}}{P_{2}}$

Substitute values in the above expression.
${[\frac{V_{2}}{24.63}]}^{1.3} = \frac{2}{5.2}$

Hence, $V_{2} = 11.81 l i t r e$
The work done is $= \frac{n R}{γ - 1} [T_{2} - T_{1}] = \frac{3 \times 8.314}{0.3} \times [250 - 200] = + 4157 J = + 4.157 k J$

Since the process is adiabatic, no heat is transferred. $q = 0$

Internal energy decreases as work is done. Hence, $Δ U = w = 4.157 k J$
Also, the expression for the enthalpy change is $Δ H = n \times C_{p} \times Δ T$
Substitute values in the above expression.
$Δ H = 3 \times 35.814 \times 50 = 5372.1 J = 5.372 k J$

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A sample of 3.0 moles of perfect gas at 200 K and 2.0 atm is compressed reversibly and adiabatically until the temperature reaches 250 K. Given that molar heat capacity is 27.5 J K−1 mol−1 at constant volume. Calculate q, w, ΔU, ΔH and the final pressure and volume.