Question

Three kg of gas in a piston–cylinder assembly undergo a process during which the relationship between pressure and specific volume is Pv0.5=constant. The process begins with P1=250 kPa and V1=1.5 m3 and ends with P2=100 kPa. Determine the final specific volume, in m3/kg.

Answer 1

The final specific volume in m³/kg is 0.0313 m³/kg.

Since $Pv^{0.5} = constant$ where P = pressure and v = specific volume, we have

Since $P_{1} v_{1}^{0.5} = P_{2} v_{2}^{0.5}$ where

• P₁ = initial pressure = 250 kPa,
• v₁ = initial specific volume = V₁/m where
• V₁ = initial volume of gas = 1.5 m³ and
• m = mass of gas = 3 kg. so, v₁ = V₁/m = 1.5 m³/3 kg = 0.5 m³/kg,
• P₂ = final pressure = 100 kPa,
• v₂ = final specific volume of gas

Final Specific volume

Making v₂ subject of the formula, we have

$v_{2} = v_{1} ( \frac{P_{1} }{P_{2} }) ^{2}$

Substituting the values of the variables into the equation, we have

$v_{2} = v_{1} ( \frac{P_{1} }{P_{2} }) ^{2}$

$v_{2} = 0.5 \frac{m^{3} }{kg} ( \frac{250 kPa }{100 kPa }) ^{2}\\v_{2} = 0.5 \frac{m^{3} }{kg} ( \frac{25}{10}) ^{2}\\v_{2} = 0.5 \frac{m^{3} }{kg} ( 0.25) ^{2}\\v_{2} = 0.5 \frac{m^{3} }{kg} (0.0625)\\v_{2} = 0.0313 \frac{m^{3} }{kg}$

The final specific volume in m³/kg is 0.0313 m³/kg.

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