A gaseous reaction occurs at a constant pressure of 35.0 and releases 71.5 of he

Question

A gaseous reaction occurs at a constant pressure of 35.0 and releases 71.5 of heat. Before the reaction, the volume of the system was 8.40 . After the reaction, the volume of the system was 2.60 . 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 is applied to the wire, the gas compresses from 5.40 to 2.70 . When the external pressure is increased to 2.50 , the gas further compresses from 2.70 to 2.16 . In a separate experiment with the same initial conditions, a pressure of 2.50 was applied to the gas, decreasing its volume from 5.40 to 2.16 in one step. If the final temperature was the same for both processes, what is the difference between for the two-step process and for the one-step process in joules?

Explanation / Answer

Delta U = q - P(Delta)V or Delta U = q + w 2.(Delta)V = 2.00 - 7.80 = -5.8 3. since (Delta)V is negative w is positive. P(delta)V = 5.8 * 40 = 232. 4. convert to kJ so 232 becomes 23.432. 5. since it release it, its loss of heat so its -71.5 + 23.432 = -48.068kJ 6. Hope its right

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