Question

A gas expands by $0.5$ litre against a constant pressure of one atmosphere. Calculate the work done in joule and calorie.
A. -50.66 J
B. -100.5 J
C. -200 J
D. -10 J

Answer 1

In this question we need to find the work done by the molecules of the gas to expand against the pressure of the constant of one atm. The work done is given by$\Rightarrow Work = - {p_{ext}} \times volume$, here p is the exerted pressure on the molecules and volume is the amount of gas which has been expanded.

Complete step by step answer:
According to the question;
The volume of the expanding gas is: $0.5$
Pressure at which gas expand is: $1$
We know that whenever an object changes its place or move from one region to another region the work is done is moving the object.
Here in this case the work is done by the gas molecules to expand themselves,
And that can be calculated using formula given in equation (1).
$\Rightarrow Work = - {p_{ext}} \times volume$
$\Rightarrow - 1 \times 0.5$
$\Rightarrow - 0.5$litres-atm
Therefore, the work done is $- 0.5$
Now we need to convert litres-atm into joule and calorie.
First we will convert it into joules;
To do so we will multiply our answer with $101.328J$
$\Rightarrow - 0.5 \times 101.328J$
$\; \Rightarrow - 50.664J$
Therefore, the work done in joule is $\; - 50.664J$
Thus the correct option is A.
Now if we want to convert it into calorie,
We will multiply our answer with $24.20$
$\Rightarrow 0.5litre - atm$
$\Rightarrow - 0.5 \times 24.20$
$\Rightarrow - 12.10cal$
Therefore, the work done in calorie is $- 12.10cal$

Note:
Joule: It is the international standard unit for the work done or the energy involved in the system, which is defined as the amount of work done when force of 1 newton is applied through the distance 1 newton in the direction of the force.
Calorie: In nutrition, the energy is calculated in terms of calorie, which is defined as the amount of heat energy which is required to raise the temperature of $1g$ of ${H_2}O$ water by one degree.