During a quasi-static expansion of a gas in an adiabatic container, the pressure

Question

During a quasi-static expansion of a gas in an adiabatic container, the pressure at any moment is given by the equation PV^gamma = K, where gamma and K are constants. Show that the work done in expanding from a state (P_i, V_i) to a state (P_f, V_f) is W = -P_i V_i - P_f V_f/gamma - 1. If the initial pressure and volume are 10^6 Pa and 10^-3 m^3, respectively, and the final values are 2 times 10^5 Pa and 3.16 times 10^-3 m^3, respectively, how much work is done on a gas having gamma = 1.4? Data Find Diagram Formulas Manipulations

Explanation / Answer

Given that

initial pressure P1=10^6 pa

volume V1=10^-3 m^3

finial pressure P2=0.2*10^6 pa

volume V2=3.16*10^-3 m^3

now we find the work done on the gas

the work done W=P2V2-P1V1/n-1

=3.16*10^-3*0.2*10^6-10^6*10^-3/1.4-1

=-0.368*10^3/0.4

=-920 J

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