III FIGURE CP19.79 shows a thermodynamic process followed by $0.015mol$of hydrogen. How much heat energy is transferred to the gas?

Amount of hydrogen $=0.015mol$

Heat given to the gas from the first law of thermodynamics as

$\Delta E=Q+W\Rightarrow \underset{¯}{Q=\Delta E-W}$

In order to change internal energy and work, we need to identify the work. Identifying the work will make it easier to solve numerically.

Typically, the work in thermodynamic processes is calculated as minus the area under the curve $pV$.

The trapezoid represents this area. F

Following the formula for the area of a trapezoid, the work will be:

$W=-\frac{(300-100)·{10}^{-6}·(4+1)·{10}^5}{2}=\underset{¯}{-50\mathrm{J}}$

Change in internal energy when the temperature changed from $T_{1}to{T}_2$is given by

$\Delta E=n{C}_v\left(T_2-T_1\right)$

From the ideal gas law, the temperatures can be given as

$pV=nRT\Rightarrow T_x=\frac{p_x{V}_x}{nR}$

Change in temperature will be

$\Delta T=\frac{p_2{V}_2-p_1{V}_1}{nR}$

So, Change in internal energy will be taken,

$\Delta E=n{C}_v\frac{p_2{V}_2-p_1{V}_1}{nR}=\frac{\left(p_2{V}_2-p_1{V}_1\right)C_v}{R}$

By Numerically,

role="math" localid="1648195809494" $\Delta E=\frac{\left({10}^5·300·{10}^{-6}-4·{10}^5·100·{10}^{-6}\right)·20.4}{8.314}=-25J$

When the amount of work and the change in internal energy are calculated, the amount of heat will be calculated,

$Q=-25+50=25\mathrm{J}$

Therefore, the amount of heat energy transferred to the gas is$25J.$