A sample of ideal gas is expanded to twice its original volume of 1 m3 in a quasi-static process for which P = \(\alpha V^2\) with \(\alpha = \) 3 x 105 Pa/m6 as shown in the given figure. Work done by the expanding gas is :
Answer & explanation
Correct answer: option 2
Work done : W = \(\int_{V_1}^{V_2} \alpha V^2 dV\)
... where \(\alpha = \) 3 x 105 Pa/m6
W = \([\frac{\alpha V^3}{3}]^2_1\)
W = 105 (V32 - V31)
V1 = 1 m3 ; V2 = 2 m3
W = 105(8-1) = 7 x 105 J