thermodynamics question

Question

400 g of air at 20°C is contained in a closed cylinder fitted with a piston The piston is allowed to expand, so that it moves with a mean force of 120N over a distance of 30mm. If the temperature drops by 0.1°C, calculate the heat transfer that must have occurred

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Answer 1

From the First Law of Thermodynamics, we know that:

ΔU = Q + W

Where Q is the heat flowing into the system, and W is the work done on the system. Rearranging for Q, you get:

Q = ΔU - W

Since the cylinder is expanding, the work is being done by the system, so the sign of W flips

Q = ΔU + W

Since the gas is air, which is mainly a mixture of nitrogen gas and oxygen gas, we can assume that it is an ideal diatomic gas. Therefore the change in internal energy is:

ΔU = (5/2) n R ΔT

The number of moles of gas, n, can be found using the molar mass of the gas, M, and the mass of the gas, m.

n = m/M

Therefore the change in internal energy is:

ΔU = (5/2)Δ(m/M) R ΔT

In thermodynamics, the work done on the gas is usually represented by the area under the pressure vs. volume graph. However, more generally, it is the area under a force versus position graph. Since we are told the mean force, that graph would be a rectangle, so:

W = F Δx

To have units of joules, the displacement would need to be measured in meters, so

Δx = 0.03 m

According to Wikipedia, the molar mass of dry air is 28.97 g/mol. The gas constant, R, is 8.31 J/(mol K). The last info needed is the change in temperature, ΔT. Since the temperature dropped:

ΔT = - 0.1 K

Putting that all together:

Q = (5/2) (m/M) R ΔT + F Δx = (5/2) [400 g / (28.97 g/mol)] ( - 0.1 K) + 120 N (0.03 m)

Q = -25 J