If a monoatomic ideal gas simultaneously expands against a constant external pressure and drops in temperature, how do you find the internal energy change?

Known values are: $T_1$, $T_2$, $p_1$, $p_2$, $p_\text{external}$ and $T_\text{external}$.

Finding the work done by the expansion is easy, but I don't understand how to change in heat.

It is not a reversible process, so it can't be found from the work, and it's not isothermic, so it isn't zero. It can be found if I have the heat capacity but that is not given.

Edit: The pressure changes from $p_1$ to $p_2 = p_\text{external}$


2 Answers 2


The internal energy $U$ is a state function that can be expressed as a function of temperature $T$, pressure $p$, and amount of substance $n$:


The change in internal energy $\Delta U$ is independent of the path by which the new state was reached. Hence, you do not necessarily have to calculate the exchanged heat $Q$ and the work $W$ done during a particular process.

The ideal gas gives simpler equations. According to Joule’s second law, the internal energy $U$ of an ideal gas depends only on the amount of gas and the temperature, and is independent of the volume or the pressure:


For a monoatomic ideal gas (i.e. a model system of point-mass molecules) we find that the internal energy is proportional to the temperature $T$ and number of molecules $N$ or amount of substance $n$:


where $k$ is the Boltzmann constant and $R$ is the molar gas constant.

Since you know the temperatures $T_1$ and $T_2$, and since the amount of substance is constant $(n_1=n_2)$, you should be able to calculate the change in internal energy $\Delta U=U_2-U_1$.

Note, if you don’t know the number of molecules $N$ or amount of substance $n$, you may express the result as molar internal energy $U_\mathrm m$ (i.e. in $\mathrm{J/mol}$):

$$U_\mathrm m=\frac Un$$


This is an adiabatic irreversible expansion so $q=0$. The change in internal energy is $\Delta U=\int_{T_1}^{T_2}C_V dT$ where $C_V$ is the heat capacity at constant volume for the monoatomic gas (=3R/2). The work done is $w=-\int_{V_1}^{V_2} pdV$ at the expense of the internal energy. As the temperature drops as no heat added or lost and then $\Delta U =w$. As heat capacity in this case is independent of temperature and the gas is ideal $\Delta U=C_V\Delta T =-p\Delta V $


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