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Latent heat of vaporisation of a liquid at 500 K and 1 atm pressure is 10.0 kcal/mol. What will be the change in internal energy (ΔE) of 3 mol of liquid at same temperature?
- a)13.0 kcal
- b)-13.0 kcal
- c)27.0 kcal
- d)-27.0 kcal
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
Latent heat of vaporisation of a liquid at 500 K and 1 atm pressure is...
Vaporization of 3 moles of H2O vapors is:
3H2O(l) → 3H2O(g)
Δn = 3 - 0 = 3
Therefore,
ΔU = ΔH - ΔnRT
= (3 x10) - 3 (0.002) (500) = 27
change in internal energy is:
ΔU = 27 Kcal
3H2O(l) → 3H2O(g)
Δn = 3 - 0 = 3
Therefore,
ΔU = ΔH - ΔnRT
= (3 x10) - 3 (0.002) (500) = 27
change in internal energy is:
ΔU = 27 Kcal
Most Upvoted Answer
Latent heat of vaporisation of a liquid at 500 K and 1 atm pressure is...
Given data:
- Latent heat of vaporisation at 500 K and 1 atm pressure = 10.0 kcal/mol
- Number of moles of liquid = 3 mol
Calculating Change in Internal Energy:
- The change in internal energy (ΔE) when a substance changes from liquid to vapor phase is given by the latent heat of vaporisation.
- Therefore, the change in internal energy for 1 mole of liquid at 500 K and 1 atm pressure is 10.0 kcal.
- For 3 moles of liquid, the total change in internal energy will be 3 times the change for 1 mole.
- ΔE = 3 * 10.0 kcal = 30.0 kcal
Final Answer:
Therefore, the change in internal energy of 3 moles of liquid at the same temperature is 30.0 kcal.
- Latent heat of vaporisation at 500 K and 1 atm pressure = 10.0 kcal/mol
- Number of moles of liquid = 3 mol
Calculating Change in Internal Energy:
- The change in internal energy (ΔE) when a substance changes from liquid to vapor phase is given by the latent heat of vaporisation.
- Therefore, the change in internal energy for 1 mole of liquid at 500 K and 1 atm pressure is 10.0 kcal.
- For 3 moles of liquid, the total change in internal energy will be 3 times the change for 1 mole.
- ΔE = 3 * 10.0 kcal = 30.0 kcal
Final Answer:
Therefore, the change in internal energy of 3 moles of liquid at the same temperature is 30.0 kcal.
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Latent heat of vaporisation of a liquid at 500 K and 1 atm pressure is 10.0 kcal/mol. What will be the change in internal energy (ΔE) of 3 mol of liquid at same temperature?a)13.0 kcalb)-13.0 kcalc)27.0 kcald)-27.0 kcalCorrect answer is option 'C'. Can you explain this answer?
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