During a reversible process, there are abstracted 317 kJ/s from 1.134 kg/s of a certain gas while the temperature remains constant at 26.7°C. For this gas, cp = 2.232 and cv = 1.713 kJ/kg.K. The initial pressure is 586 kPa. For both nonflow and steady flow (∆PE=0, ∆KE=0) process, determine (a) V1 , V2 , and P2 , (b) the work and Q, (c) ∆S and ∆H.

Answer:

Part (a) – Calculate V1, V2, and P2

For a reversible process at constant temperature, we can use the ideal gas law to relate the pressure and volume of the gas:

PV = nRT

where:

  • P is the pressure (kPa)
  • V is the volume (m³)
  • n is the number of moles of gas
  • R is the gas constant (8.314 J/mol·K)
  • T is the temperature (K)

Since the temperature is constant, we can write:

P1V1 = P2V2

We are given that P1 = 586 kPa and V1 = 1.134 m³/s. We can also calculate the number of moles of gas using the molar mass of the gas and the mass flow rate:

n = m/M

where:

  • m is the mass flow rate (kg/s)
  • M is the molar mass (kg/mol)

For a specific gas, we can find the molar mass from the specific heat capacities using the relationship:

cp - cv = R

Solving for R, we get:

R = cp - cv = 2.232 kJ/kg·K - 1.713 kJ/kg·K = 0.519 kJ/kg·K

Using the given values, we can calculate the number of moles of gas:

n = m/M = (1.134 kg/s) / (M kg/mol)

We need to find the molar mass (M) in order to proceed.

Part (b) – Calculate the work and Q

For a reversible process at constant temperature, the work done is equal to the heat transferred:

W = Q

We can calculate the work using the formula:

W = -PdV

where:

  • W is the work (kJ)
  • P is the pressure (kPa)
  • V is the volume (m³)

Since the temperature is constant, we can write:

W = -P1V1 + P2V2

Using the values we have so far, we can calculate the work:

W = (-586 kPa)(1.134 m³) + (P2)(1.134 m³) = -662.46 kPa·m³ + 1.134P2 m³

To calculate the heat transferred (Q), we can use the formula:

Q = mcΔT

where:

  • Q is the heat transferred (kJ)
  • m is the mass (kg)
  • c is the specific heat capacity (kJ/kg·K)
  • ΔT is the change in temperature (K)

Since the temperature is constant, ΔT = 0, and therefore Q = 0.

Part (c) – Calculate ∆S and ∆H

For a reversible process at constant temperature, the entropy change can be calculated using the formula:

ΔS = nR ln(P2/P1)

where:

  • ΔS is the entropy change (kJ/kg·K)
  • n is the number of moles of gas
  • R is the gas constant (kJ/mol·K)
  • P1 is the initial pressure (kPa)
  • P2 is the final pressure (kPa)

Using the values we have so far, we can calculate the entropy change:

ΔS = (n)(0.519 kJ/mol·K) ln((P2)/586 kPa)

We need to find the final pressure (P2) to complete the calculation.

The enthalpy change for a reversible process at constant temperature is zero:

ΔH = 0

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