Chemical Engineering - Chemical Engineering Thermodynamics - Discussion
Discussion Forum : Chemical Engineering Thermodynamics - Section 1 (Q.No. 19)
19.
The theoretical minimum work required to separate one mole of a liquid mixture at 1 atm, containing 50 mole % each of n- heptane and n- octane into pure compounds each at 1 atm is
Discussion:
11 comments Page 1 of 2.
AVANISH VERMA said:
5 years ago
P1 = yi.P=0.5(in mixture individual press)
P2 = P=1(in pure compound)
n = 1
n1 = 0.5
n2 = 0.5
W = -n1RTln(P1/P2) - n2RTln(P1/P2).
W = RTln(0.5).
P2 = P=1(in pure compound)
n = 1
n1 = 0.5
n2 = 0.5
W = -n1RTln(P1/P2) - n2RTln(P1/P2).
W = RTln(0.5).
Imelda Dante said:
9 years ago
by using first law of thermodynamics;
w=nRTlnV2/V1 since v2/v1 =1 and W = derivative of PV.
PV = -RTln0.5.
W = -RTln0.5.
w=nRTlnV2/V1 since v2/v1 =1 and W = derivative of PV.
PV = -RTln0.5.
W = -RTln0.5.
M.Atif said:
10 years ago
n = Total number of moles.
0.5 is individual number of mole of each component.
0.5 is individual number of mole of each component.
Mohan said:
6 years ago
But, the isothermal condition gives the maximum work, so how can it be minimum?
Bajwa said:
1 decade ago
w = -nRTlnP1/P2.
where n=1
p1 = 1.
p2 = 1+1 = 2.
=> w = -RTln 0.5.
where n=1
p1 = 1.
p2 = 1+1 = 2.
=> w = -RTln 0.5.
(1)
Prashant said:
6 years ago
@Kumar.
It's clearly given in question each have 1atm so total 2 atm.
It's clearly given in question each have 1atm so total 2 atm.
Kushal kayal said:
8 years ago
@Sunny.
Here said one-mole liquid mixture, so n=1.
Here said one-mole liquid mixture, so n=1.
Remington said:
1 decade ago
Minimum work = Helmoholtz energy.
Sunny said:
10 years ago
How can n=1? Its given as 0.5.
Gabru said:
9 years ago
How to solve this?
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