A quantity of a diatomic ideal gas undergoes a process in which both its pressur

Question

A quantity of a diatomic ideal gas undergoes a process in which both its pressure and volume are increased by a factor of n = 8 as shown in the figure below. What is the energy absorbed by heat into the gas during this process? Hint: The internal energy of a diatomic ideal gas at pressure P and occupying volume V is given by U = 5 2 PV. (Give your answer as a multiple of P0V0.) Incorrect: Your answer is incorrect. Your response differs from the correct answer by more than 10%. Double check your calculations. ? P0V0

Explanation / Answer

initial internal energy, Ui = (5/2)*Pi*Vi

= (5/2)*Po*Vo

final initenral energy, Uf = (5/2)*Pf*Vf

= (5/2)*n*Po*n*Vo

= (5/2)*Po*Vo*n^2

change in internal energy, dU = Uf - Ui

= (5/2)*Po*Vo*n^2 - (5/2)*Po*Vo

= (5/2)*Po*Vo*8^2 - (5/2)*Po*Vo

= 157.5*Po*Vo

Workdone by the gas, dW = Area under P-V curve

= Po*(nVo-vo) + 0.5*(n*Vo - Vo)*(n*Po-Po)

= Po*(8*Vo - vo) + 0.5*(8*Vo - Vo)*(8*vo - Po)

= 7*Po*Vo + 0.5*7*Vo*7*Po

= 31.5*Po*Vo

now use fisr law of thermodynamics

delta_Q = delta_U + delta_W

= 157.5*Po*Vo + 31.5*Po*Vo

= 189*Po*Vo <<<<<<<<<<------------Answer

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