Find out the relationship between boyle's temperature and van der waals constant
Answers
Answer:
This is relation between Boyle's temperature and Van der waals constant.
Explanation:
The Mathematical condition for Boyle’s temperature:
Let's See...
Mathematical condition for Boyle’s temperature is....
TB = [d(PV)/dP]T
When P → 0
and we know that...
Van der Waals Equation for 1 mole gas is...
(P + a/V²)(V - b) = RT
or,
P = (RT/V - b) - a/V²
Hence...
PV = {RTV/(V - b)} - a/V
So....
TB = [d(PV)/dP]T = [RT/(V - b) - {RTV/(V - b)²} + a/V²](dV/dP)T
= [{RT(V - b) - RTV/(V - b)²} + a/V²] (dV/dP)T
= [{- RTb/(V - b)²} + a/V²] (dV/dP)T
But when...
T = TB, [d(PV)/dP]T = 0 and (dV/dP)T ≠ 0
So...
RTBb/(V - b)² = a/V²
or,
TB = (a/Rb) {(V - b)/V}² Since P → 0, V is large.
Thus...
(V - b)/V ≃ 1
Hence...
TB = a/Rb
Answer:
Explanation:
Let's See...
Mathematical condition for Boyle’s temperature is....
TB = [d(PV)/dP]T
When P → 0
and we know that...
Van der Waals Equation for 1 mole gas is...
(P + a/V²)(V - b) = RT
or,
P = (RT/V - b) - a/V²
Hence...
PV = {RTV/(V - b)} - a/V
So....
TB = [d(PV)/dP]T = [RT/(V - b) - {RTV/(V - b)²} + a/V²](dV/dP)T
= [{RT(V - b) - RTV/(V - b)²} + a/V²] (dV/dP)T
= [{- RTb/(V - b)²} + a/V²] (dV/dP)T
But when...
T = TB, [d(PV)/dP]T = 0 and (dV/dP)T ≠ 0
So...
RTBb/(V - b)² = a/V²
or,
TB = (a/Rb) {(V - b)/V}² Since P → 0, V is large.
Thus...
(V - b)/V ≃ 1
Hence...
TB = a/Rb
This is relation between Boyle's temperature and Van der waals constant.