One kilogram of air is heated reversibly at constant pressure from an initial state of 300 K and 1 bar until its volume triples. Calculate W, Q, ΔU, and ΔH for the process. Assume air obeys the relation PV/T = 83.14 bar cm^{3} mol^{-1} K^{-1} and that C_{p} = 29 J mol^{-1} K^{-1}.

Calculations:

From the relation, PV/T = 83.14 bar cm^{3} mol^{-1} K^{-1}

Initial molal volume = 83.14 x 300 /1 = 24942 cm^{3}/mol

= 24.942 m^{3}/kmol

i.e. Initial volume of 1 kg of air (V_{1}) = 24.942 m^{3 }/ 29 = 0.86 m^{3}

Final volume (V_{2})= 3 x V_{1} = 3 x 0.86 = 2.58 m^{3} = 2.58 x 10^{6} cm^{3}.

Final specific volume = 24942 x 3 = 74826 cm^{3}/mol

Final temperature:

P_{2 }x 74826 = T_{2} x 83.14

1 x 74826 = T_{2} x 83.14

T_{2} = 900 K

For the constant pressure heating process,

W = -P(V_{2} - V_{1}) = -1 x (2.58 x 10^{6} -
0.86 x 10^{6}) bar cm^{3 }= -1 x 10^{5} x (2.58 -
0.86) Nm = -172000 J = -172 kJ. (work is done by the
system by expansion)

Moles of air in 1 kg mass = 1000/29 = 34.483 mol

Q = mC_{p}ΔT = 34.483 x 29 x (900 - 300) = 600000 J = 600 kJ.

ΔU = Q + W = 600 - 172 = 428 kJ (heat given to the system and work done on the system are positive quantities)

ΔH = Q = 600 kJ

Summary:

W |
-172 kJ |

Q |
600 kJ |

ΔU |
428 kJ |

ΔH |
600 kJ |

Last Modified on: 01-May-2024

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