Question

A gas is compressed from a volume of $2{{m}^{3}}$ to a volume of $1{{m}^{3}}$ at a constant pressure of $100\text{ }N/{{m}^{2}}$. Then it is heated at a constant volume by supplying 150 J of energy. As a result, the internal energy of the gas:
(A) Increase by 250 J
(B) Decrease by 250 J
(C) Increase by 50 J
(D) Decrease by 50 J

Answer 1

Hint: We can calculate the internal energy of the gas or system by the formula $\Delta U=Q+W$, where $\Delta U$ is the internal energy, $Q$ is the heat applied to the system, and $W$ is the work done by the system or on the system.

Complete step by step solution:
We can solve this question by the equation of the first law of thermodynamics.
The equation of the first law of thermodynamics is:
$\Delta U=Q+W$
Where $\Delta U$ is the internal energy, $Q$ is the heat applied to the system, and $W$ is the work done by the system or on the system.
Now we can calculate the work done of the or by the system as:
$W=P\Delta V$
Where P is the pressure of the gas and $\Delta V$ is the change in volume.
Given in the question, pressure of the gas is $100\text{ }N/{{m}^{2}}$and the volume changes from $2{{m}^{3}}$ to a volume of $1{{m}^{3}}$.
So the work done will be:
$W=100\text{ x }(2-1)$
$W=100\text{ J}$
And the heat applied to the system is 150 J, now applying both in the equation of first law of thermodynamics, we get
$\Delta U=150\ \text{ }+\text{ }100$
$\Delta U=250$
So, the internal energy will increase by 250 J because the value is positive.

Therefore, the correct answer is an option (A)- Increase by 250 J.

Note: The work done is not always positive. We can easily determine whether work is positive or negative. If the final volume is less than the initial volume then the work done will be positive and if the final volume is more than the initial volume then the work done will be negative.