Part B An ideal gaseous reaction (which is a hypothetical gaseous reaction that
ID: 539011 • Letter: P
Question
Part B
An ideal gaseous reaction (which is a hypothetical gaseous reaction that conforms to the laws governing gas behavior) occurs at a constant pressure of 35.0 atm and releases 71.5 kJ of heat. Before the reaction, the volume of the system was 7.40 L . After the reaction, the volume of the system was 2.00 L .
Calculate the total internal energy change, E, in kilojoules.
Express your answer with the appropriate units.
Hints
Correct
Part C
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. (Figure 2) 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 5.90 to 2.95 L . When the external pressure is increased to 2.50 atm, the gas further compresses from 2.95 to 2.36 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 5.90 to 2.36 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?
Express your answer with the appropriate units.
E = -52.4 kJExplanation / Answer
q for the two-step process = P1*V1 + P2*V2
= 2.00(5.90-2.95) Latm + 2.50(2.95-2.36) Latm
= 7.375 Latm = 7.375 Latm * 101.325 J/Latm = 747.27 J.
q for the one-step process = PV = 2.50(5.9-2.36) Latm * 101.325 J/Latm
= 896.72 J.
Hence q for the one-step process is greater by (896.72 - 747.27 ) = 149.45 J.