Question

Five moles of monatomic ideal gas have initial pressure 2.50 ×
10^{3} Pa and initial volume 2.10 m^{3}. While
undergoing an adiabatic expansion, the gas does 1180 J of work.
What is the final pressure of the gas after the expansion? in
kPa

Answer #1

work done in adiabatic process is W = K*(V2^(1-gamma)-V1^(1-gamma))/(1-gamma)

gamma value for monoatomic gas is 1.66

V1 = 2.1 m^3

P1 = 2.5*10^3 Pa

P2 = ?

V2 = ?

P2*V2^gamma = P1*V1^gamma = K = 2.5*10^3*(2.1)^1.66 = 8567

then work done is W = K*(V2^(1-gamma)-V1^(1-gamma))/(1-gamma)

1180 = 8567*((V2^(1-1.66)-2.1^(1-1.66))/(1-1.66))

V2^(-0.66) = 0.522

V2^(0.66) = 1/0.522 = 1.92

V2 = 1.92^(1/0.66) =2.68 m^3

then P2*V2^gamma = K = 8567

P2*2.68^1.66 = 8567

P2 = 1667.72 pa = 1.667 kPa

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