Figure 18-54 displays a closed cycle for a gas. From cto b, $\mathbf{40}\mathbf{}\mathbf{\text{J}}$is transferred from the gas as heat. From bto a, $\mathbf{130}\mathbf{}\mathbf{\text{J}}$is transferred from the gas as heat, and the magnitude of the work done by the gas is $\mathbf{80}\mathbf{}\mathbf{\text{J}}$. From atoc, $\mathbf{400}\mathbf{}\mathbf{\text{J}}$is transferred to the gas as heat. What is the work done by the gas from atoc? (Hint:You need to supply the plus and minus signs for the given data.)

1. The magnitude of heat transferred from the gas from c to b is,$Q_{c\to b}=-40\text{J}.$
2. The magnitude of heat transferred from the gas from b to a is,$Q_{b\to a}=-130\text{J}.$
3. The magnitude of heat transferred to the gas from a to c is,$Q_{a\to c}=400\text{J}.$
4. The magnitude of work done by the gas from b to a is,$W_{b\to a}=80\text{J}.$
5. Hint: You need to supply the plus and minus signs for the given data.

According to the First Law of Thermodynamics, energy can only be changed from one form to another and cannot be generated or destroyed. We can find the work done by the gas from a to c by applying the first law of thermodynamics to the given system.

Formula:

In an isothermal process,as per the first law of the thermodynamics (the internal energy of the system is zero), $Q_{net}=W_{net}$ …(i)

We assume that heattransferred from the gas is negative andheattransferred to the gas is positive. Also, work done by the gas is negative and work done on the gas is positive.

For thegiven isothermal process, using equation (i), we can find the work done by the path a to c as

$\begin{array}{c}{Q}_{b\to a}+Q_{a\to c}+Q_{c\to b}=W_{b\to a}+W_{a\to c}+W_{c\to b}\\ -130\text{J}+400\text{J}-40\text{J}=-80\text{J}+W_{a\to c}+0\left(\text{Sincethereisnovolumechangefromctob}\right)\\ W_{a\to c}=310\text{J}\\ =3.1\times {10}^2\text{J}\end{array}$

Therefore, the work done by the gas from a to c is$3.1\times {10}^2\text{J}.$