### Thermodynamics

# Isothermal-Isochoric Process

One mole of an ideal gas, initially at 20^{o}C 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 100^{o}C its final pressure is 10 bar. Calculate Q, W, DU, and DH for the process. Take C_{p} = 7R/2 and C_{v} = 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,

P_{1}V_{1} = RT_{1}

P_{2}V_{2} = RT_{2}

P_{3}V_{3} = RT_{3}

For the given process, T_{2} = T_{1} and V_{3} = V_{2}.

V_{1} = R x (273 + 20)/1 = 293R

And V_{3} = R x (273 + 100) / 10 = 37.3R

For the point 2, P_{2} = 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 (V_{2}/V_{1}) = 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 = C_{v}(T_{3} - T_{2}) = (5R/2) x (373 - 293) = 200R

DH = C_{p}(T_{3} - T_{2}) = (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 |