5 1) An ideal gaseous reaction (which is a hypothetical gaseous reaction that co
ID: 878187 • Letter: 5
Question
5
1) An ideal gaseous reaction (which is a hypothetical gaseous reaction that conforms to the laws governing gas behavior) occurs at a constant pressure of 50.0 atm and releases 66.7 kJ of heat. Before the reaction, the volume of the system was 7.80 L . After the reaction, the volume of the system was 2.40 L .
Calculate the total internal energy change, U, in kilojoules.
Express your answer with the appropriate units.
2) An ideal gas (which is is a hypothetical gas that conforms to the laws governing gas behavior) confined to a container with a massless piston at the top. A massless wire is attached to the piston. When an external pressure of 2.00 atm is applied to the wire, the gas compresses from 5.40 to 2.70 L . When the external pressure is increased to 2.50 atm, the gas further compresses from 2.70 to 2.16 L .
In a separate experiment with the same initial conditions, a pressure of 2.50 atm was applied to the ideal gas, decreasing its volume from 5.40 to 2.16 L in one step.
If the final temperature was the same for both processes, what is the difference between q for the two-step process and q for the one-step process in joules?
Explanation / Answer
1) dQ = 66.7 kJ
P = 50 atm
V1 =7.80 L
V2 = 2.40 L
work dW = P (V2-V1 )
= 50 (2.40-7.80)
= -270 Lit-atm (1 liter atmosphere = 101.325 joules )
= -270 x 101.235 x 10^-3
= -27.35 kJ
dQ = dU + dW
66.7 = dU -27.35
internal energy change =dU = 94.05 kJ
2)
for two step process
work = P1 (V2-V1 ) + P2 (V3-V2)
= 2 (2.70-5.40) + 2.5 (2.16-2.70)
= -5.4 -1.35
= - 6.75 Lit-atm
= -0.68 kJ
in this process temperature constant so dU =0
dQ = dW
dQ = - 0.68 kJ
for single step step process :
dQ = dW = 2.5 (2.16-5.40)
=8.1 L-atm
dQ = -0.82 kJ
difference = -0.68 +0.82
= 0.14 kJ
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