Question

A gas expands from $1.5{\text{L}}$ to $6.5{\text{L}}$ against a constant pressure of $0.5{\text{ atm}}$ and in the process the gas also absorbs 100 J of heat. The change in internal energy of the gas is:
A.$153.3{\text{ J}}$
B.$353.3{\text{ J}}$
C.$ - 153.3{\text{ J}}$
D.$ - 353.3{\text{ J}}$

Answer 1

Hint: 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:
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, $\Delta {\text{V = }}6.5{\text{ }} - {\text{ }}1.5{\text{ }} = {\text{ }}5{\text{L}}$ ; ${\text{P = }}0.5{\text{ atm}}$ ; q = 100 J
Converting these into SI units, we shall get:
$\Delta {\text{V = }}0.005{\text{ }}{{\text{m}}^{ - 3}}$ , ${\text{P = }}0.5 \times 101325{\text{ Pa}}$ , q = 100 J, substituting these values in the equation, we get:
$\Delta {\text{U = }}100{\text{ }} - {\text{ }}\left( {0.5 \times 101325 \times 0.005} \right)$
$ \Rightarrow \Delta {\text{U = }}100{\text{ }} - {\text{ }}253.31$
Solving this for internal energy, we get:
$\Delta {\text{U = }} - 153.3{\text{ J}}$

$\therefore $ The correct option is option C, i.e. $ - 153.3{\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.