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Calculate under what pressure (atm) water would boil at 150°C. If the change in specific volume when 1 gram of water is converted into steam is 1676cc. Given, latent heat of vaporization of steam = 540 cal/g; J = 4.2×107 erg/cal and one atmosphere pressure = 106 dynes/cm2
Select one:
Select one:
- a)3.628
- b)1.814
- c)2.814
- d)0.814
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
Calculate under what pressure (atm) water would boil at 150°C. If ...
Here, ∂H = 540 cal.
= 540 × 4.2 × 107 erg.
∂V = 1676cc
Boiling Temperature T = 373 K
Temperature when water is to be boil = 150°C
= 150 + 273 = 423 K.
∂T = 423 – 373 = 50 K
∂p = ?
Applying these value in the maxwell's thermodynamical relation
= 1.814 × 106 dynes/cm2
= 1.814 atmosphere
Therefore, the pressure at which water would boil at 150°C
= 1.814 + 1,000 = 2.814 atmospheric pressure.
The correct answer is: 2.814
= 540 × 4.2 × 107 erg.
∂V = 1676cc
Boiling Temperature T = 373 K
Temperature when water is to be boil = 150°C
= 150 + 273 = 423 K.
∂T = 423 – 373 = 50 K
∂p = ?
Applying these value in the maxwell's thermodynamical relation
= 1.814 × 106 dynes/cm2
= 1.814 atmosphere
Therefore, the pressure at which water would boil at 150°C
= 1.814 + 1,000 = 2.814 atmospheric pressure.
The correct answer is: 2.814
Most Upvoted Answer
Calculate under what pressure (atm) water would boil at 150°C. If ...
The boiling point of water is dependent on the pressure applied to it. At standard atmospheric pressure (1 atm), water boils at 100°C or 212°F. As the pressure decreases, the boiling point of water also decreases.
To calculate the pressure (in atm) at which water would boil at 150°C, we need to use a reference point. Let's use the boiling point of water at 100°C and 1 atm pressure.
Using the Clausius-Clapeyron equation, we have:
ln(P2/P1) = ΔHvap/R * (1/T1 - 1/T2)
Where:
P1 = 1 atm (reference pressure)
T1 = 100°C + 273.15 (reference temperature in Kelvin)
T2 = 150°C + 273.15 (desired temperature in Kelvin)
ΔHvap = heat of vaporization of water = 40.7 kJ/mol (at 100°C)
R = 8.314 J/(mol·K) (gas constant)
Plugging in the values:
ln(P2/1) = (40.7 kJ/mol) / (8.314 J/(mol·K)) * (1 / (100°C + 273.15 K) - 1 / (150°C + 273.15 K))
Simplifying:
ln(P2) = (40.7 kJ/mol) / (8.314 J/(mol·K)) * (1 / 373.15 K - 1 / 423.15 K)
ln(P2) = 4.903 - 4.580
ln(P2) = 0.323
Using the natural logarithm property:
P2 = e^0.323
P2 ≈ 1.381 atm
Therefore, water would boil at approximately 150°C under a pressure of 1.381 atm.
To calculate the pressure (in atm) at which water would boil at 150°C, we need to use a reference point. Let's use the boiling point of water at 100°C and 1 atm pressure.
Using the Clausius-Clapeyron equation, we have:
ln(P2/P1) = ΔHvap/R * (1/T1 - 1/T2)
Where:
P1 = 1 atm (reference pressure)
T1 = 100°C + 273.15 (reference temperature in Kelvin)
T2 = 150°C + 273.15 (desired temperature in Kelvin)
ΔHvap = heat of vaporization of water = 40.7 kJ/mol (at 100°C)
R = 8.314 J/(mol·K) (gas constant)
Plugging in the values:
ln(P2/1) = (40.7 kJ/mol) / (8.314 J/(mol·K)) * (1 / (100°C + 273.15 K) - 1 / (150°C + 273.15 K))
Simplifying:
ln(P2) = (40.7 kJ/mol) / (8.314 J/(mol·K)) * (1 / 373.15 K - 1 / 423.15 K)
ln(P2) = 4.903 - 4.580
ln(P2) = 0.323
Using the natural logarithm property:
P2 = e^0.323
P2 ≈ 1.381 atm
Therefore, water would boil at approximately 150°C under a pressure of 1.381 atm.
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Calculate under what pressure (atm) water would boil at 150°C. If the change in specific volume when 1 gram of water is converted into steam is 1676cc. Given, latent heat of vaporization of steam = 540 cal/g; J = 4.2×107 erg/cal and one atmosphere pressure = 106 dynes/cm2Select one:a)3.628b)1.814c)2.814d)0.814Correct answer is option 'C'. Can you explain this answer?
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