An ideal gas (which is is a hypothetical gas that conforms to the laws governing

Question

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 6.40 to 3.20 L . When the external pressure is increased to 2.50 atm, the gas further compresses from 3.20 to 2.56 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 6.40 to 2.56 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

According to first law of thermodynamics,

w = -PdV = -q

First process,

w = - 2 x (3.2 - 6.4) - 2.5 x (2.56 - 3.2) = 8.0 L.atm = 810.6 J

q = -810.6 J

Second process,

w = - 2.5 x (2.56 - 6.4) = 9.6 L.atm = 972.72 J

q = -972.72 J

Difference between q = two step - one step = -972.72 - (-810.6) = -162.12 J

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