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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 × 10) - 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 × 10) - 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...
ΔU) when 1 mole of the liquid is completely vaporised at constant temperature and pressure?
The change in internal energy (ΔU) can be calculated using the formula:
ΔU = q + w
where q is the heat absorbed or released by the system and w is the work done by or on the system.
In this case, the process is taking place at constant temperature and pressure, so there is no work done (w=0). Therefore, we can simplify the formula to:
ΔU = q
The heat absorbed (q) during the vaporisation of 1 mole of the liquid can be calculated using the formula:
q = nΔHvap
where n is the number of moles and ΔHvap is the latent heat of vaporisation.
Substituting the given values, we get:
q = 1 × 10.0 kcal/mol = 10.0 kcal/mol
Converting kcal to J (1 kcal = 4.184 kJ), we get:
q = 10.0 × 4.184 kJ/mol = 41.84 kJ/mol
Therefore, the change in internal energy (ΔU) when 1 mole of the liquid is completely vaporised at constant temperature and pressure is 41.84 kJ/mol.
The change in internal energy (ΔU) can be calculated using the formula:
ΔU = q + w
where q is the heat absorbed or released by the system and w is the work done by or on the system.
In this case, the process is taking place at constant temperature and pressure, so there is no work done (w=0). Therefore, we can simplify the formula to:
ΔU = q
The heat absorbed (q) during the vaporisation of 1 mole of the liquid can be calculated using the formula:
q = nΔHvap
where n is the number of moles and ΔHvap is the latent heat of vaporisation.
Substituting the given values, we get:
q = 1 × 10.0 kcal/mol = 10.0 kcal/mol
Converting kcal to J (1 kcal = 4.184 kJ), we get:
q = 10.0 × 4.184 kJ/mol = 41.84 kJ/mol
Therefore, the change in internal energy (ΔU) when 1 mole of the liquid is completely vaporised at constant temperature and pressure is 41.84 kJ/mol.
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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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