# Chapter 2 : Properties of thromodynamic systems¶

## Example 2.1 Page No : 24¶

In [1]:
# Variables
T = 40    			#C
R = 0.0820 			#lit-atm deg**-1 mol**-1
v = 0.381 			#lit
b = 0.043 			#lit
a = 3.6

# Calculations
P = (R*(273+T)/(v-b))-(a/v**2)

# Results
print  'Pressure  = %.1f atm'%(P)

Pressure  = 51.1 atm


## Example 2.2 Page No : 24¶

In [2]:
# Variables
T = 0 	    		#C
R = 0.0820 			#lit-atm deg**-1 mol**-1
p = 400. 			#atm

# Calculations
V = R*(273+T)/p

# Results
print  'Volume of the ideal gas  = %.4f lit mol**-1'%(V)

Volume of the ideal gas  = 0.0560 lit mol**-1


## Example 2.3 Page No : 29¶

In [1]:
# Variables
p = 400.  			#atm
T = 273 	    		#K
R = 0.0820 			#lit-atm deg**-1 mol**-1
k = 1.27

# Calculations
V = k*R*T/p

# Results
print  'Volume of the ideal gas  = %.3f lit mol**-1'%(V)

Volume of the ideal gas  = 0.071 lit mol**-1


## Example 2.4 Page No : 29¶

In [4]:
# Variables
V = 0.381 			#lit
T = 313. 			#K
R = 0.0820 			#lit-atm deg**-1 mol**-1
pc = 72.9 			#atm

# Calculations
p = R*T/V
r = p/pc

# Results
print  'Pressure of carbon dioxide gas  = %.1f atm'%(p)
print  ' ratio  = %.3f '%(r)

Pressure of carbon dioxide gas  = 67.4 atm
ratio  = 0.924


## Example 2.5 Page No : 30¶

In [5]:
# Variables
n1 = 0.25 			#mole
n2 = 0.75 			#mole
l = 0.0832 			#lit
T = 50 			#C
p1 = 404 			#atm
p2 = 390 			#atm

# Calculations
P = n1*p1+n2*p2

# Results
print  'Total Pressure  = %.f atm'%(P)

Total Pressure  = 394 atm


## Example 2.6 Page No : 31¶

In [6]:
# Variables
n1 = 0.25 			#mole
nh = 0.75 			#mole
p = 400. 			#atm
T = 50. 			#C
vn = 0.083 			#lit
vh = 0.081 			#lit

# Calculations
V = n1*vn+vh*nh

# Results
print  'Volume of given mixture is  = %.3f lit'%(V)

Volume of given mixture is  = 0.082 lit

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