Question

#5 Air undergoes an adiabatic compression in a piston-cylinder assembly from P1= 1 atm and Ti=70 oF to P2= 5 atm. Employing ideal gas model with constant specific heat capacity ratio (Y), determine the work and heat transfer per unit mass if y = 1.5. (15 points)​

Answer 1

Answer:

The work transfer per unit mass is approximately 149.89 kJ

The heat transfer for an adiabatic process = 0

Explanation:

The given information are;

P₁ = 1 atm

T₁ = 70°F = 294.2611 F

P₂ = 5 atm

γ = 1.5

Therefore, we have for adiabatic system under compression

$T_{2} = T_{1}\cdot \left (\dfrac{P_{2}}{P_{1}} \right )^{\dfrac{\gamma -1}{\gamma }}$

Therefore, we have;

$T_{2} = 294.2611 \times \left (\dfrac{5}{1} \right )^{\dfrac{1.5 -1}{1.5 }} \approx 503.179 \ K$

The p·dV work is given as follows;

$p \cdot dV = m \cdot c_v \cdot (T_2 - T_1)$

Therefore, we have;

Taking air as a diatomic gas, we have;

$C_v = \dfrac{5\times R}{2} = \dfrac{5\times 8.314}{2} = 20.785 \ J/(mol \cdot K)$

The molar mass of air = 28.97 g/mol

Therefore, we have

$c_v = \dfrac{C_v}{Molar \ mass} = \dfrac{20.785}{28.97} \approx 0.7175 \ kJ/(kg \cdot K)$

The work done per unit mass of gas is therefore;

$p \cdot dV =W = 1 \times 0.7175 \times (503.179 - 294.2611) \approx 149.89 \ kJ$

The work transfer per unit mass ≈ 149.89 kJ

The heat transfer for an adiabatic process = 0.