# physics

A sample of ideal gas is expanded to twice its original volume of 1.00 m3 in a quasi-static process for which P = áV2, with á = 2.50 atm/m6, as shown in the figure. How much work is done by the expanding gas?
i tried using the eq:
w=-1/3(2.50atm/m^6)(1.013x10^5)[(2.00m^3)-(1.00m^3)]
i end up getting 5.84x10^6 but this is not the right answer help?

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asked by tom
1. Is V2 supposed to represent V^2?

The work done by the gas equals the integral of P dV where V goes from V1 to V2 = 2V1. They tell you P as a function of V.

Just do the integration of P(V) dV. To get the answer in Joules, you will have to use P units of Newton/m^2, not atm.

W = Integral of a V^2 dV
V1 = 1.00 m^3 to V2 = 2.00 m^3

= (a/3) (V2^3 - V1^3) = (a/3)(8 - 1)m^9
where
a = 2.5*1.01*10^5 N/m^8

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posted by drwls

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