Physics 9702 · AS & A Level · The first law of thermodynamics

The first law of thermodynamics — practice question

The volume occupied by $1.00\,\text{kg}$ of water when it is a liquid at $100\,^{\circ}\text{C}$ is $1.00 \times 10^{-3}\,\text{m}^3$. For $1.00\,\text{kg}$ of water vapour at $100\,^{\circ}\text{C}$ and atmospheric pressure $1.01 \times 10^5\,\text{Pa}$, the volume is $1.69\,\text{m}^3$.
(a)[2]

Show that the work done against the atmosphere when $1.00\,\text{kg}$ of liquid water turns into water vapour is $1.71 \times 10^5\,\text{J}$.

(b(i)-1)[1]

The first law of thermodynamics may be given by the expression $\Delta U = + q + w$, where $\Delta U$ is the increase in internal energy of the system. State what is meant by $+q$.

(b(i)-2)[1]

The first law of thermodynamics may be given by the expression $\Delta U = + q + w$, where $\Delta U$ is the increase in internal energy of the system. State what is meant by $+w$.

(b(ii))[2]

The specific latent heat of vaporisation of water at $100\,^{\circ}\text{C}$ is $2.26 \times 10^6\,\text{J kg}^{-1}$. A mass of $1.00\,\text{kg}$ of liquid water becomes water vapour at $100\,^{\circ}\text{C}$. Determine, using your answer in (a), the increase in internal energy of this mass of water during vaporisation.

(b(i))[2]

The first law of thermodynamics may be given by the expression $\Delta U = +q + w$, where $\Delta U$ is the increase in internal energy of the system. State what is meant by (1) $+q$, and (2) $+w$.

Worked solution & mark scheme

This 8-mark question has a full step-by-step worked solution and mark scheme. One marking point: Volume change or acknowledgement that the liquid volume is negligible

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