In this problem, 0.95 mol of an ideal diatomic gas is heated at constant volume
ID: 1566010 • Letter: I
Question
In this problem, 0.95 mol of an ideal diatomic gas is heated at constant volume from 282 K to 546 K.
(a) Find the increase in the internal energy of the gas, the work done by the gas, and the heat absorbed by the gas.
(b) Find the same quantities if this gas is heated from 282 K to 546 K at constant pressure. Use the first law of thermodynamics and your results for Part (a) to calculate the work done by the gas.
(c) Calculate the work done in Part (b). This time calculate it by integrating the equation dW = P dV.
J
Explanation / Answer
part (a)
internal energy = n*Cv*dT
Cv = (5/2)*R
increase in internal energy dU = 0.95*(5/2)*8.314*(546-282) = 5212.8 J
since volume is constant dv = 0
Work W = P*dV = 0
heat absorbded Q = n*Cv*dT = 5212.8 J
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(b)
here pressure is constant
internal energy dU = n*Cv*dT = 5212.8 J
heat absorbed dQ = n*Cp*dT = n*(7/2)*R*dT = 7298 J
from first law
dQ = dU + W
work W = dQ - dU = 2085.2 J
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(c)
work dW = (P*dV)V)
total work W = integral dW = integral(P*dV)
W = P*(V2-V1)
W = P*V2 - P*V1
W = n*R*T2 - n*R*T1
W = n*R*(T2-T1)
W = 0.95*8.314*(546-282)
W = 2085.1 J
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