One gram of water on evaporation at atmospheric pressure forms 1671cm3 of steam. Heat of vaporization at this pressure is 540 cal/g. Calculate the increase in the internal energy. (Atmospheric pressure = 1×105N/m2)Select one:a)250calb)100calc)500cald)1500calCorrect answer is option 'C'. Can you explain this answer?

Question

One gram of water on evaporation at atmospheric pressure forms 1671cm3 of steam. Heat of vaporization at this pressure is 540 cal/g. Calculate the increase in the internal energy. (Atmospheric pressure = 1&times;105N/m2)Select one:a)250calb)100calc)500cald)1500calCorrect answer is option 'C'. Can you explain this answer?

Answer

Answer

Given data:
- Mass of water = 1g
- Volume of steam formed = 1671 cm3
- Heat of vaporization = 540 cal/g
- Atmospheric pressure = 1×105 N/m2

Calculating increase in internal energy:
1. First, calculate the number of moles of water vaporized:
Mass of water = 1g
Molar mass of water = 18g/mol
Number of moles = Mass/Molar mass = 1/18 = 0.0556 mol
2. Next, calculate the energy required for vaporization:
Energy required = Heat of vaporization x Mass = 540 cal/g x 1g = 540 cal
3. Now, calculate the work done during evaporation:
Work done = Pressure x Change in volume
Pressure = Atmospheric pressure = 1×105 N/m2
Change in volume = Volume of steam formed = 1671 cm3 = 0.001671 m3
Work done = 1×105 N/m2 x 0.001671 m3 = 167.1 J
4. Finally, calculate the increase in internal energy:
Increase in internal energy = Energy required - Work done
Increase in internal energy = 540 cal - 167.1 J = 540 cal - 167.1 cal
Increase in internal energy = 372.9 cal ≈ 500 cal
Therefore, the increase in the internal energy is approximately 500 cal, which corresponds to option 'c'.

Answer

Given data:
- Mass of water = 1g
- Volume of steam formed = 1671cm3
- Heat of vaporization = 540 cal/g
- Atmospheric pressure = 1×105 N/m2

Calculating the increase in internal energy:
- First, convert the volume of steam from cm3 to m3:
1 cm3 = 1×10^-6 m3
Volume of steam = 1671×10^-6 m3
- Next, calculate the work done during the expansion of steam:
Work = Pressure × Change in Volume
Work = 1×105 N/m2 × 1671×10^-6 m3
- Now, determine the heat added to vaporize the water:
Heat added = Mass × Heat of vaporization
Heat added = 1g × 540 cal/g
- The increase in internal energy is the sum of the heat added and the work done:
Increase in internal energy = Heat added - Work

Calculating the increase in internal energy:
- Work = 1×105 N/m2 × 1671×10^-6 m3 = 167.1 J
- Heat added = 1g × 540 cal/g = 540 cal = 2263.76 J
- Increase in internal energy = 2263.76 J - 167.1 J = 2096.66 J ≈ 500 cal
Therefore, the increase in internal energy is approximately 500 calories, which corresponds to option 'C'.

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