To a closed system 150 kJ of work is supplied. If the initial volume is 0.6 m³ and pressure of the system changes as p = 8 – 4V, where p is in bar and V is in m³, determine the final volume and pressure of the system.

Question

Accepted Answer

Amount of work supplied to a closed system = 150 kJ

Initial volume = 0.6 [latex]m ^3[/latex]

Pressure-volume relationship, p = 8 – 4V

The work done during the process is given by

[latex]W=\int_{V_1}^{V_2} p d V[/latex]

[latex]=10^5 \int_{0.6}^{V_2}(8-4 V) d V=10^5\left[8 V-4 imes \frac{V^2}{2} ight]_{0.6}^V[/latex]

[latex]=10^5\left[8\left(V_2-0.6 ight)-2\left(V_2^2-0.6^2 ight) ight][/latex]

[latex]=10^5\left[8 V_2-4.8-2 V_2^2+0.72 ight][/latex]

[latex]=10^5\left[8 V_2-2 V_2^2-4.08 ight][/latex] Nm or J

But this work is equal to [latex]-150 imes 10^3[/latex] J as this work is supplied to the system.

[latex] herefore \quad-150 imes 10^3=10^5\left[8 V_2-2 V_2^2-4.08 ight][/latex]

or [latex]2 V_2^2-8 V_2+2.58=0[/latex]

[latex]V_2=\frac{8 \pm \sqrt{64-4 imes 2 imes 2.58}}{4}=\frac{8 \pm 6.585}{4}=0.354[/latex] [latex]m ^3[/latex]

Positive sign is incompatible with the present problem, therefore it is not considered.

∴ Final volume, [latex]V_2= 0 . 3 5 4[/latex] [latex]m ^3[/latex].

and, final pressure, [latex]p_2=8-4 V=8-4 imes 0.354[/latex]

[latex]=6.584 ext { bar }=6.584 imes 10^5[/latex] [latex]N / m ^2[/latex] or Pa.

Question 2.20: To a closed system 150 kJ of work is supplied. If the initia...

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