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Igoryamba
2 years ago
10

The virial equation of state for n-butane at 460 k is z 1 a/v b/v2 in which a 265 cm3 /g mole and b 30,250 cm6 /g mole2. calcula

te the work required
Health
1 answer:
const2013 [10]2 years ago
4 0

Answer:

- 1384.7 J.

Explanation:

STEP ONE: make use of the equation of state to determine the amount of pressure.

The equation of state is given below as;

NRT = PV. Where P = pressure, V = volume, R = gas constant, T= temperature and N = number of moles.

Therefore, making P the subject of the formula, we have;

P = NRT/V.

Thus, P = (1 + a/V + b/V^2) [(RT)/V].

Note that a = -265 cm^3/g = - 0.265 m^3/kg and b = 30250 cm^6/g = 0.030250 m^6/kg.

Hence, P = ( 1 - 265/V + 0.030250/V^2) {(RT)/V}.

P = ( 1/V - 265/V^2 + 0.030250/V^3) × RT.

STEP TWO: Calculate for the volume using the equation of state in step one above.

V = ( 1 - 265/V+ 0.030250/V^2) × nRT/P

V = ( 1 - 265/V+ 0.030250/V^2) × (1 × 0.08205 × 460/50) = 0.39L.

Thus, we have that the V above = V1 = 0.39L.

Also, V = ( 1 - 265/V+ 0.030250/V^2) × (1 × 0.08205 × 460/100) = 0.1769L.

Workdone, W = Integration of - PdV.

Taking V1 and V2 as the lower and upper limits.

W = - ∫ PdV = - ∫ ( 1/V - 265/V^2 + 0.030250/V^3) dV.

W = - 37.7 + (-0.82237 + 0.8264 - 0.3862). = 13.67 × 101325 × 10^-3 = - 1384.7J.

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