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in Physics by (65 points)
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0.014 m3 gas at a pressure of 2070 kn/mexpands to a pressure of 207kn/maccording to the law pv1.35 = c. determine the work done by the gas during expansion.

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1 Answer

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\(V_1 = 0.014\)

\(P_1 = 2070\, kn/m^2\)

\(P_2 = 207\, kn/m^2\)

\(V_2 = ?\)

\(v = 1.35\)

we know

\(V_2 = V_1(\frac{P_1}{P_2})^{1/r}\)

\(V_2 = 0.014 \left(\frac{2070}{207}\right)^{1/1.35}\)

\(V_2 = 0.014 \times (10)^{0.74}\)

\(V_2 = 0.076 m^3\)

Work done 

\(W =\frac1{v-1} (P_1V_1 - P_2V_2)\)

\(= \frac1{(1.35-1)} (2070 \times 0.014 - 207 \times 0.076)\)

\(= \frac 1{0.35} (28.98 - 15.73)\)

\(= \frac1{0.35} \times 13.25\)

\(W = 37.85 J\)

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