Q1: 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

ANS:B - 0.065

To solve this problem, we can use the fact that the volume of the vessel remains constant (rigid walled vessel). We'll use the volume of the vessel and the volumes of the steam and water vapor at different temperatures to determine the amount of water vapor condensed. Given:

• Initial temperature (T1) = 100°C
• Initial pressure (P1) = 1.01325 bar
• Initial volume of steam (V1) = 1.673 m³
• Final temperature (T2) = 98°C
• Final pressure (P2) = 0.943 bar
• Volume of water vapor at final conditions (V2) = 1.789 m³

We know that the volume of the rigid vessel remains constant. Therefore, the initial volume of steam (V1) is equal to the final volume of water vapor (V2) plus the volume of condensed water vapor (V_condensed). So, we have: 𝑉1=𝑉2+𝑉condensedV1​=V2​+Vcondensed​ 1.673 m3=1.789 m3+𝑉condensed1.673m3=1.789m3+Vcondensed​ 𝑉condensed=1.673 m3−1.789 m3Vcondensed​=1.673m3−1.789m3 𝑉condensed=−0.116 m3Vcondensed​=−0.116m3 Since the volume can't be negative, it means no water vapor condensed. So, the amount of water vapor condensed is 0 kg. Therefore, the correct answer is 0.0 kg.