Question
- What is an adiabatic process? Derive expression for the work done during such process.
- A refrigerator is to maintain eatable kept inside at 9°C. If the room temperature is 36°C. Calculate the co-efficient of performance.
Derivation: Consider one gram mole of an ideal gas enclosed in a cylinder with perfectly non-conducting walls, and fitted with a perfectly frictionless, non-conducting piston. P1 = Let be the initial pressure, v1 be the volume and T1 be the temperature of the gas.
Force exerted by the gas on the piston of area of cross-section ‘A’ is given by
F = P × A ...(i)
Where P is the pressure at any instant during expansion.
If we assume that pressure of the gas during an infinitesimally small outward displacement dx of the piston remains constant, then small amount of work done during expansion,
dW = F × dx = (P × A)dx
dW = PdV ...(ii)
Where dV = A(dx) = small increase in the volume of the gas. Total work done by the gas in adiabatic expansion from volume v1 to v2 is
$\text{W}=\int\limits^{\text{V}_2}_{\text{V}_1}\text{P dV}\dots\text{(iii)}$
The equation of adiabatic changes is
PV' = K, a constant ...(iv)
Where, $\gamma=\frac{\text{C}_\text{P}}{\text{C}_\text{V}}$
$=\frac{\text{Specific heat of gas at constant pressure}}{\text{Specific heat of gas at constant volume}}$
For equation (iv)
$\text{P}=\frac{\text{K}}{\text{V}^\gamma}=\text{KV}^{-\gamma},$ put in equation (iii)
$\text{W}=\int\limits^{\text{V}_2}_{\text{v}_1}\text{KV}^{-\gamma}\text{dV}=\text{K}\Big[\frac{\text{V}^{(1-\gamma)}}{1-\gamma}\Big]^{\text{v}_2}_{\text{v}_1}$
$\text{W}=\frac{\text{K}}{1-\gamma}\Big[\text{K}\nu^{(1-\gamma)}_2-\text{K}\nu^{(1-\gamma)}_1\Big]$
$\text{W}=\frac{1}{1-\gamma}\Big[\text{K}\nu^{(1-\gamma)}_2-\text{K}\nu^{(1-\gamma)}_1\Big]\dots\text{(v)}$
From standard gas equation P1 V1 = RT, P2V2 = RT2
$\text{W}=\frac{1}{1-\gamma}[\text{RT}_2-\text{RT}_1]$
[Putting above in equation (v)]
$\text{W}=\frac{\text{R}(\text{T}_2-\text{T}_1)}{1-\gamma}$
T1 = 36°C = 36 + 273 = 309K
Coeffiecient of performance
$=\frac{\text{T}_2}{\text{T}_1-\text{T}_2}=\frac{282}{309-282}=\frac{282}{27}=10.4$
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