In: Physics
A cylinder with a piston contains 0.200 mol of nitrogen at 1.70×105 Pa and 320 K . The nitrogen may be treated as an ideal gas. The gas is first compressed isobarically to half its original volume. It then expands adiabatically back to its original volume, and finally it is heated isochorically to its original pressure.
Part A
Find the work done by the gas during the initial compression.
W = |
-266 |
J |
Correct
Part B
Find the heat added to the gas during the initial compression.
Q = | -930 | J |
Correct
Part C
Find internal-energy change of the gas during the initial compression.
ΔU = | -664 | J |
Correct
Part D
Find the work done by the gas during the adiabatic expansion.
Enter your answers numerically separated by commas.
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W = ? | J |
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Part E
Find the heat added to the gas during the adiabatic expansion.
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Q = | ? | J |
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Part F
Find the internal-energy change of the gas during the adiabatic expansion.
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ΔU = | ? J |
(D)
Use the following formula,
PV = nRT
(1.7x 105 pa) V = (0.2) (8.31) (320)
V = 0.00313 m3
During compression,
V1= initial volume = 0.00313 m3
V2= final volume = V1/2 = 0.00313/2 m3 = 0.00156 m3
T1 = initial temperature = 320 K
Use the formula,
V1/T1 = V2/T2
0.00313 /320 = 1.56 x 10-3 /T2
T2 = 150 K
During adiabatic expansion,
P1 = initial pressure = 1.7 x 105 pa
V1 = initial Volume = 1.56 x 10-3
V2= final volume = 0.00313
P2 = final pressure
For adiabatic process
P1 V1 = P2 V2
For diatomic gas nitrogen, = 1.4
(1.7 x 105 ) (1.56 x 10-3 )1.4 = P2 (0.00313)1.4
P2 = 6.44 x 104 pa
The work done is given as,
W = (P2 V2 - P1 V1) / (1 - )
W = ((6.44 x 104 pa ) (0.00313) - (1.7 x 105 ) (1.56 x 10-3 )) / (1-1.4)
W = 158.93 J
(E)
Heat added = Q =0
(F)
From the first law of thermodynamics,
Q = U + W
U = Q - W = 0 – 158.93 J
U = -158.93 J