Isothermal-Isochoric Process

One mole of an ideal gas, initially at 20oC and 1 bar, undergoes the following mechanically reversible changes. It is compressed isothermally to a point such that when it is heated at constant volume to 100oC its final pressure is 10 bar. Calculate Q, W, DU, and DH for the process. Take Cp = 7R/2 and Cv = 5R/2.

Calculations:

The given process is represented in PV diagram as follows:

We can find the pressure and volume of points 1, 2 and 3 by equation of state (ideal gas equation)

For one mole of ideal gas,

P1V1 = RT1

P2V2 = RT2

P3V3 = RT3

For the given process, T2 = T1 and V3 = V2.

V1 = R x (273 + 20)/1 = 293R

And V3 = R x (273 + 100) / 10 = 37.3R

For the point 2, P2 = R x (273 + 20) / (37.3R) = 293/37.3 = 7.855 bar

PVT conditions of the points 1,2 and 3's are:

Point

P, bar

V

T, Kelvin

1

1

293R

293

2

7.855

37.3R

293

3

10

37.3R

373

 

For the isothermal process:

DU = 0

DH = 0

W = ò PdV = ò RT dV/V = RT ln (V2/V1) = R x 293 x ln (37.3/293) = -604R

Q = -W = 604R (since Q + W = DU)

Work is done on the system and heat is given out by the system

For the constant volume heating (isochoric process):

W = 0

DU = Cv(T3 - T2) = (5R/2) x (373 - 293) = 200R

DH = Cp(T3 - T2) = (7R/2) x (373 - 293) = 280R

Q = DU - W = 200R

Summary:

Process

Q (heat given to the system)

W (work done on the system)

DU

DH

Isothermal compression

-604R = -604 x 8.314 = -5021.66 J

604R = 5021.66 J

0 J

0 J

Constant volume heating

200R = 1662.8 J

0 J

1662.8 J

280R = 2327.92 J