An ideal gaseous reaction (which is a hypothetical gaseous reaction that conform

Question

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 69.5 kJ of heat. Before the reaction, the volume of the system was 8.00 L . After the reaction, the volume of the system was 2.20 L .

Calculate the total internal energy change, ?E, in kilojoules.

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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 4.40 to 2.20 L . When the external pressure is increased to 2.50 atm, the gas further compresses from 2.20 to 1.76 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 4.40 to 1.76 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. ????

help with this 2 part problem1!!!!

Explanation / Answer

I am solving the problem 1 as per Chegg guidelines, post multiple problems to get the remaining answers

Q1)

Initial volume = 8.00L

Final Volume = 2.20L

Work Done = -Pext * (V2-V1) = -50(2.20-8.00) = 290 atm.L

1 atm.L = 101.325J

Work Done = 290 (101.325J) = 29384.25J = 29.384 kJ

Change in Internal energy = work done - heat = 29.384 - 69.5 = -40.116 kJ

If significant figures are mentioned in the problem, then use -40.1 or -40.2 kJ

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