Question

1.5 kg of a gas expands within an elastic container from 100 kPaa and 150 liters to a final pressure of 8 kPaa. If the process is defined by pV1.25 = C and the internal energy decreases by 30 kJ/kg, calculate the heat transferred in kJ

Answer 1

Answer:

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Answer 2

Given info : 1.5 kg of a gas expands within an elastic container from 100 kPa and 150 liters to a final pressure of 8 kPa. If the process is defined by pV1.25 = C and the internal energy decreases by 30 kJ/kg.

To find : The heat transferred in kJ.

solution : given $PV^{1.25}=C$

so, $P_1V_1^{1.25}=P_2V_2^{1.25}$

here, P₁ = 100 kPa , V₁ = 150 Litres, P₂= 8 kPa and V₂ = ?

so, 100 × (150)^(1.25) = 8 × V₂^(1.25)

⇒12.5 × (150)^(1.25) = V₂^(1.25)

⇒(12.5)^(1.25) × 150 = V₂

⇒V₂ = 3525.5654 Litres ≈ 3525.6 litres

using formula, work done in quasi - static process is given by, $W=\frac{P_1V_1-P_2V_2}{n-1}$ when $PV^n=C$

so, Workdone , W= (100 × 150 - 3525.6 × 8)/(1.25 - 1)

= -13204.8/0.25

= -52819.2 kPa . litre

= -52819.2 × 10¯³ kJ

= -52.8192 kJ ≈ -53 kJ

internal energy change, ∆U = ∆U/∆m × m

= 30 kJ/kg × 1.5 kg

= 45 kJ

by first law of thermodynamics,

Q = ∆U + W

⇒Q = 45 kJ + (-53 kJ) = -8 kJ

therefore the heat transferred from system is 8 kJ.