Chemical Engineering - Chemical Engineering Thermodynamics - Discussion
Discussion Forum : Chemical Engineering Thermodynamics - Section 2 (Q.No. 28)
28.
Grams of butane (C4H10) formed by the liquefaction of 448 litres of the gas (measured at (STP) would be
Discussion:
8 comments Page 1 of 1.
Mohammed said:
2 years ago
One mole of grams of stp occupies 22.4l.
Amodh said:
4 years ago
22.4 is a gms mole of any gas.
Karan parmar said:
6 years ago
58/22.4 = 2.58,
2.58*448 = 1160.
2.58*448 = 1160.
(3)
Rudra said:
7 years ago
Nice explanation. Thanks.
Dilipkumar said:
8 years ago
Lets take ideal gas equation.
pv=nrt.
We know n = no of moles = weight(w)/molecular weight(m)
substitute n = w/m in ideal gas equation.
at stp T= 273k.
P= 101325.
given v = 448 liters = 0.448m3.
R = 8.314.
m = molecular weight of butane 58,
substitute all above values,
(101325*0.448*58) / (8.314*273).
We get w = 1159.9 gm.
pv=nrt.
We know n = no of moles = weight(w)/molecular weight(m)
substitute n = w/m in ideal gas equation.
at stp T= 273k.
P= 101325.
given v = 448 liters = 0.448m3.
R = 8.314.
m = molecular weight of butane 58,
substitute all above values,
(101325*0.448*58) / (8.314*273).
We get w = 1159.9 gm.
(3)
Ankeeta said:
8 years ago
Because At STP, one mole of gas occupies 22.4 L of volume.
Ravi said:
8 years ago
Why we divide by 22.4? Please explain.
Anonymous said:
9 years ago
448/22.4 = 20 moles.
20 * 58 (m.Wt) = 1160.
20 * 58 (m.Wt) = 1160.
(2)
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