Question

The pressure-volume correlation for a non-flow quasi-static process is given by P= (8-4V) b, where p is in bar, where V is in m3. If 140 kJ of work in supplied to the system, determine the final pressure and volume of the system. Take the initial volume as 0.6m3.

Answer 1

Amount of work supplied to a closed system = 150 kJ

Initial volume 0.6 m²

Pressure-volume relationship, p = 8 - 4V

The work done during the process is given by;

$W =\int^{V_2}_{V_1} pdV$

$=10⁵\int^{V_2}_{0.6} (8-4V)dV$

$=10⁵[8V-4×\dfrac{V²}2]^V_{0.6}$

$= 10⁵ [8(V_2-0.6)-2(V_2²-0.6²)]\\ = 10⁵[8V_2-4.8-2V_2² +0.72]\\ = 10 [8V_2-2V_2²-4.08] J$

But this work is equal to 150 x 10³ J as this work is supplied to the system

$\therefore\quad -150 × 10^3= 10⁵(8V_2-2V_2²-4.08)\\ 2V_{2} ^ 2 - 8V_{2} + 2.58 = 0$

$V₂ =\dfrac{ 8± \sqrt{64-4×2×2.58}}4= \dfrac{ 8±6.585}4 = 0.354 m²$

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

Therefore, Final Volume (V₂) = 0.354m³

Final pressure (P₂) = $8-4V$ = $8-(4 ×0.354)$ = 6.584 bar.