For a temperature increase of$\mathit{\Delta }{\mathit{T}}_{\mathbf{1}}$, a certain amount of an ideal gas requires$\mathbf{30}\mathbf{\text{\hspace{0.17em}J}}\mathbf{}$ when heated at constant volume and $\mathbf{50}\mathbf{\text{\hspace{0.17em}J}}\mathbf{}$when heated at constant pressure. How much work is done by the gas in the second situation?

The table containing four situations for an ideal gas is given.

1. Heat required by the gas in situation (a),$Q_V=30\text{\hspace{0.17em}J}$
2. Heat required by the gas in situation (b),$Q_p=50\text{\hspace{0.17em}J}$.

Using the relation between change in energy and change in temperature, we can equate the change in energy for both situations. Next, using the first law of thermodynamics, we can write it in terms of heat energy and work done. Using the given values and value of work done at a constant volume, we can find the work done by the gas in the second situation.

Formulae:

According to first law of thermodynamics, the change in internal energy is given by

$\Delta E=Q-W$…(i)

The change in internal energy at constant volume$\Delta E=n{C}_V\Delta T$,.…(ii)

In both cases, the change in temperature$\Delta {\text{T}}_1$is constant.

Thus, using the condition in equation (ii), we get that the change in internal energy in the two cases is also the same.

$\Delta E_V=\Delta E_p$ …(iii)

whererepresent the change in internal energy atconstant volume and pressure respectively.

At constant volume, work done is zero. So

$W_V=0$

Now, using equation (i) in equation (iii), we can get that the work done in the second situation is given as follows:

$\begin{array}{c}{Q}_V-W_V=Q_p-W_p\\ 30\text{\hspace{0.17em}J}-0\text{\hspace{0.17em}J}=50\text{\hspace{0.17em}J}-{\text{W}}_p\\ {\text{W}}_p=20\text{\hspace{0.17em}J}\end{array}$

Therefore, the work done by gas in the second situation is $20\text{\hspace{0.17em}J}$.