Consider a cylinder fitted with a piston that contains 2 mol

Consider a cylinder fitted with a piston that contains 2 mol of H_2O in a container at 1000 K. Calculate how much work is required to isothermally and reversibly compress this gas from 10 L to 1 L, in each of the following cases: Use the Ideal gas model for water. Use the Redlich-Kwong equation to relate P, v, and T Use the Steam tables.

Solution

a)

Ideal gas equation, Pv = nRT

P = nRT/v

Work done = Integral P dv

= Integral (nRT/v) dv

For isothermal process, T remains constant. Hence,

W = nRT Integral (1/v) dv

W = nRT ln (v2 / v1)

= 2* 8.314 * 1000 ln (1/10)........R is universal gas constant = 8.314 J / mol-K

= -38287.28 J

b)

Redlich Kwong equation

P = RT / (V-b) - a / (T^0.5V(V+b))

Here V is molar volume = Volume / n.

V1 = 1/2 = 0.5 L/mol = 0.0005 m³/mol

V2 = 10/2 = 5 L/mol = 0.005 m³/mol

W = Integral P dv

W = Integral [RT / (V-b) - a / (T^0.5V(V+b))] dv

For constant T we get,

W = RT ln [(V2 -b) / (V1-b)] - (a / T^0.5) (1/b) [ln (V2 / (V2+b)) - ln (V1 / (V1 + b))]

For water, a = 9.869 Pa-K^0.5-m⁶ / mol² and b = 21.1*10^-6 m³ / mol

Using R = 8.314 J/ mol-K

Putting values,

W = -18917.93 J/mol

For 2 moles, we get

W = -18917.93*2

W = -37835.86 J

c)

For water, gas constant R = 461.89 J/kg-K, Molecular mass M = 18

Mass m = 2 moles water = 18*2 g water = 36 g water = 0.036 kg water

Initial specific volume = V1 / m = 10 / 0.036 L / kg = 0.01 / 0.036 m³/kg = 0.278 m³/kg

Final specific volume = V2 / m = 0.001 / 0.036 m³/kg = 0.0278 m³/kg

Using steam tables,

From steam tables, at 1000 K and v = 0.0278 m³/kg we get u1 = 3460 kJ/kg

From steam tables, at 1000 K and v = 0.278 m³/kg we get u2 = 3520 kJ/kg

q - w = u2 - u1

0 - w = 3520 - 3460

w = -60 kJ/kg

For 0.036 kg water,

W = 0.036*(-60)

W = -2.16 kJ = -21600 J