Question

A gas is compressed at constant pressure $50N/{m^2}$ from a volume of $10{m^3}$ to a volume of $4{m^3}$. Energy $100J$ is then added to the gas by heating. Its internal energy is:
A) Increased by $100J$
B) Increased by $200J$
C) Decreased by $100J$
D) Increased by $400J$

Answer 1

Hint: A work is done on the system of gas to compress its volume. This work is required to exert a macroscopical force on the system. In a constant external pressure, the compression of gas leads to a compression of volume. This increases the frequency of the collisions. Eventually, the energy of the system increases.

Formulae Used:
As the first law of thermodynamics implies, the change in internal energy $\Delta U$ of a system can be defined as
$\Delta U = Q - \Delta W$
where $Q$ is the total heat energy of the system and the $\Delta W$ is the net work done.
For a constant external pressure, the change in volume of a gas from ${V_{initial}}$ to ${V_{final}}$, the work is done can be calculated as
$\Delta W = P \times \left( {{V_{final}} - {V_{initial}}} \right)$

Complete step by step answer:
Given:
The gas is compressed at a constant external pressure $50N/{m^2}$.
Initial volume of the gas is ${V_{initial}} = 10{m^3}$.
The final volume of the gas is ${V_{final}} = 4{m^3}$.
The added heat energy to the system is $Q = 100J$.
We need to find a change in the internal energy of the system.

Step 1:
The gas is compressed at a constant external pressure.
Calculate the net work done from eq (2) by using the values given.
$\Delta W = P \times \left( {{V_{final}} - {V_{initial}}} \right) $
$ = 50 \times \left( {4 - 10} \right)N.m $
$ = 50 \times \left( { - 6} \right)J $
$ = - 300J $

Step 2:
Now, you have calculated the net work done $\Delta W$ and you have the total heat energy added to the system $Q = 100J$.
Calculate the change in internal energy $\Delta U$
$\Delta U = Q - \Delta W$
$ \Rightarrow \Delta U = 100J - \left( { - 300J} \right)$
$ = 100J + 300J $
$ = 400J $
So the internal energy has a positive increment of $400J$ that is the internal energy is increased with $400J$.

If a gas is compressed at constant pressure $50N/{m^2}$ from a volume of $10{m^3}$ to a volume $4{m^3}$. Energy $100J$ is then added to the gas by heating, then its internal energy is Increased by $400J$. So, option (D) is correct.

Note:
The compression decreases the volume making the frequency of the collisions to be increased. This eventually increases the energy of the system. Hence you get negative work done. This negative signature is very crucial. You need to calculate the change in volume from the final to initial. Hence you shall get the correct value of net work done which is negative clearly indicating a work done on the system.