# Internal energy calculation

The temperature of 1 mole of a liquid is raised by heating it with 750 joules of energy. It expands and does 200 joules of work, calculate the change in internal energy of the liquid.

I want to use the expression: $$\Delta U = \Delta Q + \Delta W$$ so that: $$\Delta U = 750\,\mathrm J- 200\,\mathrm J = 550\,\mathrm J$$

but it strikes me that it can't be that simple (first year college exam paper). Also, of what significance is the '1 mole' of liquid?

• You proposed the right solution. Nothing to do with the amount of matter nor agregation state. – user1420303 Mar 24 '16 at 11:47
• Yes. Couldn't leave it at that though, a comment has to be longer than three characters. "1 mole of liquid" is of no significance. – getafix Mar 24 '16 at 13:43
• It is $Q$ and $W$ not $\Delta Q$ or $\Delta W$ – orthocresol Mar 24 '16 at 16:40

Your calculation is correct. The standardized definition of the change in internal energy $U$ for a closed thermodynamic system is

$$\Delta U=Q+W$$

where $Q$ is amount of heat transferred to the system and $W$ is work done on the system (provided that no chemical reactions occur). Therefore, heat transferred to the system is assigned a positive sign in the equation $$Q=750\ \mathrm J$$ whereas work done by the system on the surroundings during the expansion of the liquid is assigned a negative sign $$W=-200\ \mathrm J$$ Thus, the change in internal energy is \begin{align} \Delta U&=Q+W\\ &=750\ \mathrm J-200\ \mathrm J\\ &=550\ \mathrm J\\ \end{align}

However, the question is a bit flawed since the given values are not typical for a liquid. By way of comparison, realistic values for water are shown in the following table.

$$\textbf{Water (liquid)}\\ \begin{array}{lllll} \hline \text{Quantity} & \text{Symbol} & \text{Initial value (0)} & \text{Final value (1)} & \text{Change}\ (\Delta) \\ \hline \text{Amount of substance} & n & 1.00000\ \mathrm{mol} & 1.00000\ \mathrm{mol} & 0\\ \text{Volume} & V & 18.0476\ \mathrm{ml} & 18.0938\ \mathrm{ml} & 0.0462\ \mathrm{ml} \\ & & 1.80476\times10^{-5}\ \mathrm{m^3} & 1.80938\times10^{-5}\ \mathrm{m^3} & 4.62\times10^{-8}\ \mathrm{m^3} \\ \text{Pressure} & p & 1.00000\ \mathrm{bar} & 1.00000\ \mathrm{bar} & 0 \\ & & 100\,000\ \mathrm{Pa} & 100\,000\ \mathrm{Pa} & 0 \\ \text{Temperature} & T & 20.0000\ \mathrm{^\circ C} & 29.9560\ \mathrm{^\circ C} & 9.9560\ \mathrm{^\circ C} \\ & & 293.1500\ \mathrm{K} & 303.1060\ \mathrm{K} & 9.9560\ \mathrm{K} \\ \text{Internal energy} & U & 1\,511.59\ \mathrm{J} & 2\,261.58\ \mathrm{J} & 749.99\ \mathrm{J} \\ \text{Enthalpy} & H & 1\,513.39\ \mathrm{J} & 2\,263.39\ \mathrm{J} & 750.00\ \mathrm{J} \\ \hline \end{array}$$

When $1\ \mathrm{mol}$ of water with an initial temperature of $T_0=20\ \mathrm{^\circ C}$ is heated with $\Delta H=Q=750\ \mathrm J$ at a constant pressure of $p=1\ \mathrm{bar}$, the resulting expansion is actually only \begin{align} \Delta V&=V_1-V_0\\ &=18.0938\ \mathrm{ml}-18.0476\ \mathrm{ml}\\ &=0.0462\ \mathrm{ml}\\ &=4.62\times10^{-8}\ \mathrm{m^3} \end{align}

The corresponding pressure-volume work is \begin{align} W&=p\Delta V\\ &=100\,000\ \mathrm{Pa}\times4.62\times10^{-8}\ \mathrm{m^3}\\ &=0.00462\ \mathrm J \end{align} which is clearly below the value given in the question $(W=200\ \mathrm J)$.

The values given in the question are appropriate for a gas. For example, realistic values for nitrogen are shown in the following table.

$$\textbf{Nitrogen (gas)}\\ \begin{array}{lllll} \hline \text{Quantity} & \text{Symbol} & \text{Initial value (0)} & \text{Final value (1)} & \text{Change}\ (\Delta) \\ \hline \text{Amount of substance} & n & 1.00000\ \mathrm{mol} & 1.00000\ \mathrm{mol} & 0\\ \text{Volume} & V & 24.3681\ \mathrm{l} & 26.5104\ \mathrm{l} & 2.1423\ \mathrm{l} \\ & & 0.0243681\ \mathrm{m^3} & 0.0265104\ \mathrm{m^3} & 0.0021423\ \mathrm{m^3} \\ \text{Pressure} & p & 1.00000\ \mathrm{bar} & 1.00000\ \mathrm{bar} & 0 \\ & & 100\,000\ \mathrm{Pa} & 100\,000\ \mathrm{Pa} & 0 \\ \text{Temperature} & T & 20.0000\ \mathrm{^\circ C} & 45.7088\ \mathrm{^\circ C} & 25.7088\ \mathrm{^\circ C} \\ & & 293.1500\ \mathrm{K} & 318.8588\ \mathrm{K} & 25.7088\ \mathrm{K} \\ \text{Internal energy} & U & 6\,081.06\ \mathrm{J} & 6\,616.83\ \mathrm{J} & 535.77\ \mathrm{J} \\ \text{Enthalpy} & H & 8\,517.87\ \mathrm{J} & 9\,267.87\ \mathrm{J} & 750.00\ \mathrm{J} \\ \hline \end{array}$$

When $1\ \mathrm{mol}$ of nitrogen with an initial temperature of $T_0=20\ \mathrm{^\circ C}$ is heated with $\Delta H=Q=750\ \mathrm J$ at a constant pressure of $p=1\ \mathrm{bar}$, the resulting pressure-volume work is

\begin{align} W&=p\Delta V\\ &=100\,000\ \mathrm{Pa}\times0.0021423\ \mathrm{m^3}\\ &=214.23\ \mathrm{J} \end{align} The corresponding enthalpy balance \begin{align} \Delta H&=\Delta U+W\\ 750.00\ \mathrm{J}&=535.77\ \mathrm{J}+214.23\ \mathrm{J} \end{align} is quite similar to the values of the question $(\Delta H=Q=750\ \mathrm J,$ $\Delta U=550\ \mathrm J,$ and $W=200\ \mathrm{J}).$