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dlinn [17]
3 years ago
11

One kilogram of saturated steam at 373 K and 1.01325 bar is contained in a rigid walled vessel. It has a volume of 1.673 m3. It

is cooled to a temperature at which the specific volume of water vapour is 1.789 m. The amount of water vapour condensed in kilograms is (a) 0.0 (b) 0.065 (c) 0.1 (d) 1.0
Chemistry
1 answer:
NeTakaya3 years ago
8 0

Answer: Option (b) is the correct answer.

Explanation:

The given data is as follows.

        Initial volume (v_{1}) = 1.673 m^{3}

          Final volume (v_{2}) = 1.789 m^{3}

As, the amount of water vapor condensed will be as follows.

                     \frac{(v_{2} - v_{1})}{v_{2}}

                     = \frac{(1.789 m^{3} - 1.673 m^{3})}{1.789 m^{3}}

                     = \frac{0.116 m^{3}}{1.789 m^{3}}

                     = 0.065 kg

Hence, we can conclude that the amount of water vapour condensed in kilograms is 0.065 kg.

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The Calvin cycle, also called the light-independent or carbon fixation reactions, is the second stage of photosynthesis where water, and carbon dioxide (CO2) from air, are converted into organic compounds (i.e. sugars) using the energy from short-lived electronically excited carriers (ATP and NADPH) for the reactions. These organic compounds can then be used by the producing organism (i.e. plants) and the animals that feed on it. 

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                       6 total G3P molecules * 8 sets of 3-turns  = 48 G3P molecules


The figure below shows the products of the cycle after 3 turns (Source: https://ka-perseus-images.s3.amazonaws.com/2f4bdc8f8275834d3f5ef434d93bf16b991b2357.png). 

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