Chemical Engineering - Stoichiometry - Discussion
Discussion Forum : Stoichiometry - Section 4 (Q.No. 7)
7.
A gas at 0°C was subjected to constant pressure cooling until its volume became half the original volume. The temperature of the gas at this stage will be
Discussion:
2 comments Page 1 of 1.
Chary said:
4 years ago
V1/T1 =V2/T2.
V1 is 1 let V2 be 0.5 . And T1 is 273k. T2=?
1/273=0.5/T2.
T2=136.5k.
T2= 136.5k-273 = -136.5C.
V1 is 1 let V2 be 0.5 . And T1 is 273k. T2=?
1/273=0.5/T2.
T2=136.5k.
T2= 136.5k-273 = -136.5C.
(1)
Imelda D. said:
9 years ago
Assume 1 mole of gas; PV=nRT.
Since V2=1/2 V1;P(1/2)=nRT: T= P(22.4L/mole)(1/2)/0,08205L-atm/mole-K.
T = 136.5k-273 = -136.5C.
Since V2=1/2 V1;P(1/2)=nRT: T= P(22.4L/mole)(1/2)/0,08205L-atm/mole-K.
T = 136.5k-273 = -136.5C.
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