Question

At one bar pressure, the volume of a gas is $0.6{\text{ litres}}$ . If the gas receives $122{\text{ Joules}}$ of heat at one atmosphere pressure, the volume becomes $2{\text{ litres}}$ , then calculates its internal energy. $\left( {1{\text{ litre bar}} = 101.32{\text{ Joules}}} \right)$

Answer 1

To answer this question, you must recall the first law of thermodynamics. The first law of thermodynamics states that the energy of the universe in total remains constant always but it may undergo transformation from one form to the other.

Formula used: $\Delta U = q + W$
Where, $\Delta U$ is the internal energy of the system
$q$ is the amount of heat supplied
And, $W$ is the amount of work done by the system.

Complete step by step solution:
The first law of thermodynamics states that, Energy can neither be created nor be destroyed. It just converts from one form to the other. In other words, the total energy of the universe does not change. From this, we can also infer that in an isolated system, the energy of the system is constant.
The equation of the first law gives a relation between the internal energy of the system, heat supplied to it and the work done by the system.
In the question we are given, the heat supplied to the system, the change in volume and we need to find the change in internal energy of the system. So using the first law, we can write
$\Delta U = q + W = q + P\Delta V$
Substituting the values, we get, $\Delta U = 122{\text{J}} + (1)(2 - 0.6){\text{L}}{\text{.bar}}$
$\Delta U = 122J + \left( {101.32 \times 1.4} \right)J$
Simplifying:
$\Rightarrow \Delta U = 122 + 141.848$
$\therefore \Delta U = 263.848{\text{ Joules}}$ .

Note:
According to the first law of thermodynamics, on cyclic transformation subjected in any system, the sum of heat and work done over the process is zero. Heat supplied and work done are both path functions whereas internal energy is a state function. The cyclic integration of heat and work done over the process is zero. And since in a cyclic process, the initial and final state of the system is the same, the internal energy will also be zero. Thus, this can be used as a proof for the equation, $\Delta U = q + W$ .