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Question: A sample of an ideal gas is expanded 1 m³ to 3 m³ in a reversible process for which P = KV², with K ...

A sample of an ideal gas is expanded 1 m³ to 3 m³ in a reversible process for which P = KV², with K = 6 bar/m⁶. The magnitude of work done by the gas is

A

70 kJ

B

5200 kJ

C

15.6 kJ

D

38.6 kJ

Answer

5200 kJ

Explanation

Solution

The work done by a gas in a reversible process is given by the integral: W=−∫V1V2P dVW = -\int_{V_1}^{V_2} P \, dV

The process is defined by the relation P=KV2P = KV^2, where K=6 bar/m6K = 6 \, \text{bar/m}^6. The initial volume is V1=1 m3V_1 = 1 \, \text{m}^3 and the final volume is V2=3 m3V_2 = 3 \, \text{m}^3.

Substitute the expression for PP into the work integral: W=−∫V1V2(KV2) dVW = -\int_{V_1}^{V_2} (KV^2) \, dV

Since KK is a constant, it can be taken out of the integral: W=−K∫V1V2V2 dVW = -K \int_{V_1}^{V_2} V^2 \, dV

Now, perform the integration: W=−K[V33]V1V2W = -K \left[ \frac{V^3}{3} \right]_{V_1}^{V_2}

Substitute the limits of integration: W=−K(V233−V133)W = -K \left( \frac{V_2^3}{3} - \frac{V_1^3}{3} \right) W=−K3(V23−V13)W = -\frac{K}{3} (V_2^3 - V_1^3)

Now, substitute the given values: K=6 bar/m6K = 6 \, \text{bar/m}^6 V1=1 m3V_1 = 1 \, \text{m}^3 V2=3 m3V_2 = 3 \, \text{m}^3

W=−6 bar/m63((3 m3)3−(1 m3)3)W = -\frac{6 \, \text{bar/m}^6}{3} \left( (3 \, \text{m}^3)^3 - (1 \, \text{m}^3)^3 \right) W=−2 bar/m6(27 m6−1 m6)W = -2 \, \text{bar/m}^6 \left( 27 \, \text{m}^6 - 1 \, \text{m}^6 \right) W=−2 bar/m6(26 m6)W = -2 \, \text{bar/m}^6 \left( 26 \, \text{m}^6 \right)

Multiplying the numerical values and units: W=−(2×26) (barm6×m6)W = -(2 \times 26) \, \left( \frac{\text{bar}}{\text{m}^6} \times \text{m}^6 \right) W=−52 bar⋅m3W = -52 \, \text{bar} \cdot \text{m}^3

To express the work done in Joules, we use the conversion factor: 1 bar = 10510^5 Pa 1 m³ = 1 m31 \, \text{m}^3 Therefore, 1 bar⋅\cdotm³ = 10510^5 Pa⋅\cdotm³ = 10510^5 J.

Convert the work done to Joules: W=−52×105 JW = -52 \times 10^5 \, \text{J} W=−5,200,000 JW = -5,200,000 \, \text{J}

To express the work done in kilojoules (kJ), divide by 1000: W=−5200 kJW = -5200 \, \text{kJ}

This is the work done on the gas. The question asks for the magnitude of the work done by the gas. Work done by the gas = −W-W. Work done by the gas = −(−5200 kJ)=5200 kJ-(-5200 \, \text{kJ}) = 5200 \, \text{kJ}.

The magnitude of the work done by the gas is 5200 kJ.