Question

A sample of ideal gas ($\gamma =1.4$) is heated at constant pressure. If an amount $140J$ of heat is supplied to the gas, find
a) The change in internal energy of the gas
b) The work done by the gas.

Answer 1

The first law of thermodynamics is solely based on the concept of law of conservation of energy. Change in internal energy can be calculated by using the value of heat and $\gamma ~$by deriving a direct relationship between the two from the existing equation of change in internal energy.
the value of change in internal energy, can help us calculate the value of work done by the system, this is because the sum of heat and work done by the system, gives the value of change in internal energy.
Formula used: ${{C}_{p}}/{{C}_{V}}=\gamma ~$
Where ${{C}_{p}}$ is the heat of capacity at constant pressure and $~{{C}_{v}}$ is the heat of capacity at constant volume, and $\gamma ~$ is the ratio of specific heats

Complete step by step answer:
a) Constant pressure and Constant volume heat capacities are very important in the calculation of many changes occurring in a system. The ratio ${{C}_{p}}/{{C}_{V}}=\gamma ~$ appears in many physical expressions as well.
According to the conditions given in the question, it is well established that the heat is supplied to the system. Therefore, the value of heat and $\gamma ~$ is the ratio of specific heats that is, ${{C}_{p}}$and$~{{C}_{v}}$
$q=140J$,$\gamma =1.4$
Where $\gamma ~$ is the ratio of specific heats, and $q$ denotes the value of quantity of heat.
As we know that, the first law of thermodynamics tells us about the law of conservation of energy. Meaning, the energy can neither be created nor destroyed, it can only be transformed from one form to another. So this law can be expressed by establishing a relationship between the internal energy of the system and the specific $heat$ of the system at constant volume, along with temperature of the system. That is,
$\Delta U=n{{C}_{v}}dT$
Here, $\Delta U$ denotes the change in internal energy of the system. Internal energy of a system refers to all the energy which is present in a given system, including the kinetic energy of molecules and the energy stored in all of the chemical bonds present between molecules. With the interactions of heat, work and internal energy, there are energy transfers and conversions every time a change is made upon a system.

${{C}_{v}}$ Denotes the specific heat, at constant volume
$n$ Denotes the number of moles
And $dT$ denotes small temperature at which the gas is being heated
In the next step we will simply divide and multiply the right hand side of the equation with ${{C}_{p}}$ in order to get a ratio of ${{C}_{p}}$ and ${{C}_{v}}$as it will give the value of $\gamma$, so,
$~\Rightarrow \Delta U=\dfrac{{{C}_{v}}}{{{C}_{p}}}\times \left( {{C}_{p}}ndT \right)$
as we can see the above equation is a substituted version of the change in internal energy which was given earlier. In the later one we multiplied and divided the whole left side of the equation with the same factor so the outcome remains unchanged.
now we will replace the ratio of ${{C}_{p}}$ and ${{C}_{v}}$ with $\gamma$, and the value of $~\left( {{C}_{p}}ndT \right)$ with $q$as the heat
$\Rightarrow \Delta U=\dfrac{1}{\gamma }\times q\text{ }\left( \because \dfrac{{{C}_{v}}}{{{C}_{p}}}=\gamma \text{ }and\text{ }q=n{{C}_{p}}dT \right)$
now we put the values of as per the question,
$\therefore \Delta U=\dfrac{q}{\gamma }=\dfrac{140}{1.4}=100J$
We got the value of internal energy as of the gas $100J$
b) for the second part of the question,
Now if we consider the first law of thermodynamics,
$\Delta U=q+w$
where $w$ is the work done
$\Rightarrow w=\Delta U-q$
the value of change in internal energy is $100$ and the value of heat exchange is $140$
now we will substitute the values given in the above equation
$\Rightarrow w=100-140=-40J$
Here the presence of negative signs only indicates that the work is done by the gas.
Hence the work done by the gas is $40J$.

Note:
When the value of work is negative, then it means that work is done by the system because when the system does the work, it loses its energy and the volume of the system increases. If work is positive, then the work is being done on the system because the system is absorbing energy and the volume decreases.