Air at 1 bar and 298.15 K (25 degree C) is compressed to 5 bar and 298.15 K by t

Question

Air at 1 bar and 298.15 K (25 degree C) is compressed to 5 bar and 298.15 K by two different mechanically reversible processes: Cooling at constant pressure followed by heating at constant volume. Heating and constant volume followed by cooling at constant pressure. Calculate the heat and work requirements and Delta U and Delta H of the air for each path. The following heat capacities for air may be assumed independent of temperature: C_V = 20.78 J/(mol K) and C_P = 29.19 J/(mol K) Assume also for air PV/T is a constant, regardless of the changes it undergoes. At 298.15 K and 1 bar the molar volume of air is 0.02479 m^3/mol.

Explanation / Answer

(a)

First calculate the final volume of gas as follows:

V2 = V1* P1/P2= 0.02479 m^3 * (1/5)= 4.958*10^-3 m^3

at constant pressure the temperature of gas will be T2

T2 = T1*V2/V1 = 298.15 * 4.958*10^-3 m^3 / 0.02479 m^3 = 59.63 K

Now calcualte the dH as follows:

Q = dH = Cp * (T2-T1)=29.19 J (mol / K) (59.63 K -298.15 K)= -6962.29 J/mol

Now calcaulte the dU as follows:

dU = dH – PdV = -6962.29 J/mol – 1*10^5 (4.958*10^-3 m^3 - 0.02479 m^3)

=- 4956.45J

(b)

Q= dU = Cv* (T2-T1)=20.78 J (mol / K) (59.63 K -298.15 K)= -4956.45 J/mol

Now calculate the complete energy in this process:

-6962.29 J/mol+ 4956.45 J/mol = -2005.84 J/ mol

dU = 0

First law of thermodynamics:

dU = Q+W

0= --2005.84 J/ mol+W

W= +2005.84 J/ mol

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