Question

An ideal gas at (P,V,T) is expanding adiabatically to 5.66 times the volume and half the temperature. The degree of freedom f of the gas and work done W by the gas –

• A. W = 0, f = 5
• B. W = PV, f = 7
• C. W =$\frac{25}{23}$P, f = 7
• D. W =$\frac{25}{23}$PV, f = 5

Answer 1

Answer: (D)

W =$\frac{25}{23}$PV, f = 5

Answer 2

For an adiabatic process

T₁$V_{1}^{\gamma - 1}$ = T₂$V_{2}^{\gamma - 1}$

Given T₂ =$\frac{T_{1}}{2}$ and

V₂ = 5.66 V₁

$\frac{T_{1}}{T_{2}}$=$\left( \frac{V_{2}}{V_{1}} \right)^{\gamma - 1}$

2 = (5.66)^g–1

or g = 1.458

Hence gas is diatomic having f = 5.

W =$\frac{P_{2}V_{2} - P_{1}V_{1}}{r - 1}$ and$\frac{P_{2}V_{2}}{T_{2}}$=$\frac{P_{1}V_{1}}{T_{1}}$

=$\frac{\frac{5.66}{11.32}P_{1}V_{1} - P_{1}V_{1}}{1.46 - 1}$

P₂ =$\frac{P_{1}}{11.32}$ and

V₂ = 5.66 V₁

=$\frac{0.5P_{1}V_{1}}{0.46}$=$\left\lbrack \frac{25PV}{23} \right\rbrack$