Question
- Define isothermal process. Derive an expression for work done during isothermal process.
- A refrigerator is to remove heat from the eatables kept inside at 10°C. Calculate the coefficient of performance if room temperature is 36°C.
✓
Answer
- Isothermal process is a change in pressure and volume of a gas without any change in its temperature. In such a change, there is a free exchange of heat between the gas and its surroundings.
For a small change in volume, work done is given by
dW = PdV
We know, PV = nRT
$\therefore\text{P}=\frac{\text{nRT}}{\text{V}}$
For T = constant,
$\text{dW}=\text{nRT}\frac{\text{dV}}{\text{V}}$
Net work done under isothermal condition to change the volume from Vi to Vf is,
$\text{W}=\int\limits^{\text{V}_\text{f}}_{\text{V}_\text{i}}\text{dW}=\text{nRT}\int\limits^{\text{V}_\text{f}}_{\text{V}_\text{i}}\frac{\text{dV}}{\text{V}}$
$=\text{nRT}\big|\log_\text{e}\text{V}\big|^{\text{V}_\text{f}}_{\text{V}_\text{i}}$
$\text{W}=\text{nRT }\log_\text{e}\Big(\frac{\text{V}_\text{f}}{\text{V}_\text{i}}\Big)$
$\therefore\text{W}=2.3026\text{ nRT }\log_\text{10}\Big(\frac{\text{V}_\text{f}}{\text{V}_\text{i}}\Big)$
Where n is the number of moles. If Pf and Pi are the pressures, we can also write,
$\text{W}=2.3026\ \text{nRT }\log_{10}\Big(\frac{\text{P}_\text{i}}{\text{P}_\text{f}}\Big)$
- Coefficient of performance, $\beta=\frac{\text{T}_2}{\text{T}_1-\text{T}_2}$
$\because\text{T}_1=36 +273=309\text{K}$
and $\text{T}_2=10+273=283\text{K}$
$\therefore\beta=\frac{283}{309-283}=\frac{283}{26}=10.8$
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