Question

# An ideal gas (which is a hypothetical gas that conforms to the laws governing gas behavior)...

An ideal gas (which is a hypothetical gas that conforms to the laws governing gas behavior) is confined in a container with a massless piston at the top. 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 5.40 to 2.70L. When the external pressure is increased to 2.50 atm, the gas further compresses from 2.70 to 2.16 L.

In a seperate experiment with the same initial conditions, a pressure of 2.50 atm was applied to the ideal gas, decreasing its volume from 5.40 to 2.16 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?

We know that, the internal energy is a state function. The initial and final states of the system are identical in both cases, i.e, change in internal energy (ΔU) is the same for both cases.

From the first law of thermodynamics, we have, ΔU = q + w

where q is the heat added to the system and w is the work done on the system by the surroundings.

If only expansion work is done, w = - P * dV, where P is the constant external pressure and dV is the volume change.

In the first case, the work is done on the system in two steps:

w1 = -(2 atm)*(2.70 L - 5.40 L) - (2.50 atm)*(2.16 - 2.70 L)

i.e, w1 = 6.75 atm*L

1 atm*L = 101.325, so

Therefore, w1 = 683.94 J

Now, In the second case,

w2 = -(2.50 atm)*(2.16 L - 5.40 L) * (101.325 J/(atm*L))

i.e, w2 = 820.732 J

Equating the change in energy for the two cases, we get,

q1 + 683.94 J = q2 + 820.732 J

Therefore, the difference between q for the two-step process and q for the one-step process in joules, Δq = q1 - q2 = (820.732 - 683.94 ) J = 136.792 J

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