2023 A/L Physics Past Paper · Question 10 (English Medium)

Answer: (1) only by (A).

What the question asks

We want: in which of the processes (A), (B) and (C) the whole of the heat supplied to an ideal gas becomes work done by the gas.

What each letter means:

$\Delta Q$ ($\text{J}$): the heat supplied to the gas.

$\Delta U$ ($\text{J}$): the change of the internal energy of the gas.

$\Delta W$ ($\text{J}$): the work done by the gas.

$p$ ($\text{Pa}$): the pressure of the gas.

$\Delta V$ ($\text{m}^3$): the change of the volume of the gas.

$T$ ($\text{K}$): the absolute temperature of the gas.

Relevant theory

For the isothermal process ∆θ = 0 as temperature is constant) ∴ KE = 0 (as KE ∝ θ in kelvin) ∴ ΔU = 0 (for an ideal gas internal energy is totally kinetic energy) i.e. Internal energy of the system is constant. The 1st law ∆Q = ∆U + ∆W Now becomes, ∆Q = ∆W If ∆Q is + ve, then ΔW will also be + ve (i.e. work is done by the system).

Pressure on the piston = p ∴ Force on the piston = p.A Work done by the gas = p.A.x = p.Ax = p.∆V

The volume of a fixed mass of a gas is directly proportional to its absolute temperature, provided the pressure remains constant.

Reading the options

The options are combinations of (A), (B) and (C).

We test each process on its own.

For example, think of pocket money. You get some money, that is the heat. You can spend it on a gift, that is the work. Or you can keep it in your purse, that is the internal energy.

If the money in your purse stays the same, you must have spent the whole of what you got.

$$\Delta Q = \Delta U + \Delta W$$

In words: the heat supplied goes into the internal energy and into the work.

So the whole of the heat becomes work only when $\Delta U = 0$.

Reasoning and calculation

In the isothermal process, the whole of the heat supplied is converted to work done by the gas.

In the isothermal process the temperature stays the same. The internal energy of an ideal gas is all kinetic energy and goes with the temperature, so $\Delta U = 0$. The first law gives $\Delta Q = \Delta W$.

A

In the constant volume process, the whole of the heat supplied is converted to work done by the gas.

In the constant volume process $\Delta V = 0$, so $\Delta W = p\,\Delta V = 0$. The gas does no work. The heat goes into the internal energy, and the gas gets warmer.

B

In the constant pressure process, the whole of the heat supplied is converted to work done by the gas.

In the constant pressure process the gas expands, and its volume goes with $T$. So $T$ goes up and $\Delta U$ is not zero. Part of the heat raises the internal energy.

C

Statement (A) is true, because the internal energy does not change.

Statement (B) is false, because the gas does no work.

Statement (C) is false, because part of the heat warms the gas.

So only (A) is true.

Why the other options fail

In (B) the volume does not change, so the gas does no work. The heat goes into the internal energy.

In (C) the volume grows, and the volume goes with the absolute temperature. So the gas gets warmer and part of the heat is not work.

In (C) the gas gets warmer, so part of the heat raises the internal energy. (A) is true but (C) is false.

In (B) there is no work at all, and in (C) part of the heat warms the gas. Only (A) is true.

Conclusion

Only in the isothermal process (A) does the whole of the heat become work. (B) and (C) are false.

The answer is (1).

Write $\Delta Q = \Delta U + \Delta W$ and ask where the heat goes.

The whole of the heat becomes work only when the internal energy does not change. For an ideal gas that means the temperature does not change.

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