Question

A gas expands from 2L to 6L against a constant pressure of $0.5{\text{ atm}}$ on absorbing 200 J of heat, calculate the change in internal energy.

Answer 1

To solve this question, it is required to have knowledge about the first law of thermodynamics. The heat supplied to the system is related to internal energy (formula given). We shall substitute the given values in the equation to find out the change in internal energy.

Formula used: $\Delta {\text{U = q - W}}$ where q is the heat, $\Delta {\text{U}}$ is the change in internal energy, P is the pressure of the gas and W is the work done.

Complete step by step answer:
As we know that, according to the first law of thermodynamics, the change in internal pressure is equal to the sum of heat supplied to the system and the work done.
${\text{q = }}\Delta {\text{U + W}}$ (Eq. 1)
Now, work done in a gaseous system is defined as the change in the temperature and pressure. It is written as:
${\text{W = }}\Delta \left( {{\text{PV}}} \right){\text{ = P}}\Delta {\text{V + V}}\Delta {\text{P}}$
In the question, as the pressure is constant, $\Delta {\text{P = 0}}$ , so the equation gets converted to:
${\text{W = P}}\Delta {\text{V}}$
Putting this value in Eq. 1 we get:
$\Delta {\text{U = q - P}}\Delta {\text{V}}$
In the given question, $\Delta {\text{V = 6 - 2 = 4L}}$; ${\text{P = }}0.5{\text{ atm}}$ ; q = 200 J
Converting these into SI units, we shall get:
$\Delta {\text{V = 0}}{\text{.004 }}{{\text{m}}^{ - 3}}$ , ${\text{P = 0}}{\text{.5}} \times {\text{101325 Pa}}$ , q = 200 J, substituting these values in the equation, we get:
$\Delta {\text{U = 200 - }}\left( {0.5 \times 101325 \times 0.004} \right)$
$\Rightarrow \Delta {\text{U = 200 - 202}}{\text{.65}}$
Solving this for internal energy, we get:
$\Delta {\text{U = - 2}}{\text{.65 J}}$
$\therefore$ The change in internal energy is $- 2.65{\text{ J}}$ .

Note: In the equation to calculate change in enthalpy of a reaction, $\Delta {\text{H = }}\Delta {\text{U - P}}\Delta {\text{V}}$ , the enthalpy change can also regarded as the heat absorbed by the system. Thus, $\Delta {\text{H}}$ will be equal to heat.