Part A: A gaseous reaction occurs at a constant pressure of 45.0 a t m and relea

Question

Part A:

A gaseous reaction occurs at a constant pressure of 45.0atm and releases 67.0kJ of heat. Before the reaction, the volume of the system was 6.60L . After the reaction, the volume of the system was 2.20L .

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

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Part B:

A gas is 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 2.95L . When the external pressure is increased to 2.50 atm, the gas further compresses from 2.95 to 2.36L .

In a separate experiment with the same initial conditions, a pressure of 2.50 atm was applied to the gas, decreasing its volume from 5.90 to 2.36L 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

Delta U = q - P(Delta)V or Delta U = q + w
(Delta)V = 2.20 - 6.60 = -4.4
since (Delta)V is negative w is positive. P(delta)V = 4.4 * 45 = 198 J = 0.198 kj

since it release it, its loss of heat so its -67 + 0.198 = -66.802kJ

b)

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