Five moles of gas initially at a pressure of 2.00 atm and a volume of 0.300 L ha
ID: 2243576 • Letter: F
Question
Five moles of gas initially at a pressure of 2.00 atm and a volume of 0.300 L has internal energy equal to 79.0 J. In its final state, the gas is at a pressure of 1.50 atm and a volume of 0.800 L, and its internal energy equals 168 J.
WIAF =The correct answer is not zero. J WIBF =
Your response differs from the correct answer by more than 10%. Double check your calculations. J WIF =
Your response differs from the correct answer by more than 10%. Double check your calculations. J Five moles of gas initially at a pressure of 2.00 atm and a volume of 0.300 L has internal energy equal to 79.0 J. In its final state, the gas is at a pressure of 1.50 atm and a volume of 0.800 L, and its internal energy equals 168 J. For the paths IAF, IBF, and IF in the figure above, calculate the work done on the gas. For the paths IAF, IBF, and IF in the figure above, calculate the net energy transferred to the gas by heat in the process.
Explanation / Answer
1 atm = 101.325 kPa
a)
W_iaf = W_ia + W_af
But W_ia = 0 (since constant volume process) and
W_af = P*(Vf - Vi) = 1.5*101.325*(0.8 - 0.3) = 75.994 J
Hence, W_iaf = 75.994 J
W_ibf = W_ib + W_bf
But W_bf = 0 (since constant volume process) and
W_ib = P*(Vb - Vi) = (2*101.325)*(0.8 - 0.3) = 101.325 J
Hence, W_ibf = 101.325 J
W_if = Area under if and x-axis
= [1/2*(2+1.5)*(0.8 - 0.3)] *101.325
= 88.659 J
b)
Q_iaf = W_iaf + U_iaf
= 75.994 + (168 - 79)
= 164.994 J
Q_ibf = W_ibf + U_ibf
= 101.325 + (168-79)
= 190.325 J
Q_if = W_if + U_iaf
= 88.659 + (168-79)
= 177.659 J
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