Chemical Engineering - Stoichiometry - Discussion
Discussion Forum : Stoichiometry - Section 7 (Q.No. 1)
1.
One kg of saturated steam at 100°C and 1.01325 bar is contained in a rigid walled vessel. It lias a volume of 1.673 m3. It cools to 98°C ; the saturation pressure is 0.943 bar ; one kg of water vapour under these conditions has a volume of 1.789 m3. The amount of water vapour condensed (in kg) is
Discussion:
2 comments Page 1 of 1.
Maha said:
4 years ago
Yes, I think the correct option is D.
Maha said:
4 years ago
P1 = 1.01325 bar, V1 = 1.673 m3, T1 = 373 K.
P2 = 0.943 bar, V2 = ?, T2 = 371 K,
(P1*V1)/T1 = (P2*V2)/T2,
V2 = (P1*V1*T2)/(T1*P2).
V2 = 1.788 m3.
One kg of water vapour under these conditions has a volume of 1.789 m3.
As V2 calculated is 1.788 m3, so the amount of water vapour condensed is 1 kg.
P2 = 0.943 bar, V2 = ?, T2 = 371 K,
(P1*V1)/T1 = (P2*V2)/T2,
V2 = (P1*V1*T2)/(T1*P2).
V2 = 1.788 m3.
One kg of water vapour under these conditions has a volume of 1.789 m3.
As V2 calculated is 1.788 m3, so the amount of water vapour condensed is 1 kg.
(2)
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