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To change the melting point of ice by 1K, what should be the change in the pressure, given that L = 80 cal gm-1 and specific volume of ice and water at 0°C are 1 cm3 and 1.091 cm3 respectively. Give your answer in atm.
Correct answer is '135.25'. Can you explain this answer?
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Verified Answer
To change the melting point of ice by 1K, what should be the change in...
dT = 1K
dp = 1.3525 × 108 dynes/cm2
⇒dp = 135.25 atm.
The correct answer is: 135.25
Most Upvoted Answer
To change the melting point of ice by 1K, what should be the change in...
To change the melting point of ice by 1K, we need to use the Clausius-Clapeyron equation, which relates the change in melting point to the change in pressure. The equation is given by:
ln(P2/P1) = ΔHvap/R * (1/T1 - 1/T2)
Where:
P1 and P2 are the initial and final pressures,
ΔHvap is the enthalpy of vaporization,
R is the gas constant,
T1 and T2 are the initial and final temperatures.
In this case, we want to change the melting point of ice by 1K, so T1 is the initial temperature of ice (0°C or 273K), and T2 is the final temperature of ice (1°C or 274K).
We also know that the enthalpy of vaporization of water is 80 cal/gm.
Now, we need to find the change in pressure (P2 - P1) to change the melting point of ice by 1K. Let's assume P1 is the atmospheric pressure (1 atm or 760 mmHg).
ln(P2/1 atm) = (80 cal/gm)/(R) * (1/273K - 1/274K)
To convert the specific volume from cal/gm to J/kg, we multiply by 4184 J/cal.
ln(P2/1 atm) = (80 cal/gm * 4184 J/cal)/(R * 273K * 1000 gm/kg) * (1/273K - 1/274K)
Simplifying the equation:
ln(P2/1 atm) = (80 * 4184)/(273 * R * 1000) * (274 - 273)/(273 * 274)
Now, we can solve for P2:
P2/1 atm = e^((80 * 4184)/(273 * R * 1000) * (274 - 273)/(273 * 274))
Using the ideal gas constant R = 8.314 J/mol·K:
P2/1 atm = e^((80 * 4184)/(273 * 8.314 * 1000) * (274 - 273)/(273 * 274))
P2 ≈ 1.0003 atm
Therefore, to change the melting point of ice by 1K, the change in pressure should be approximately 0.0003 atm.
ln(P2/P1) = ΔHvap/R * (1/T1 - 1/T2)
Where:
P1 and P2 are the initial and final pressures,
ΔHvap is the enthalpy of vaporization,
R is the gas constant,
T1 and T2 are the initial and final temperatures.
In this case, we want to change the melting point of ice by 1K, so T1 is the initial temperature of ice (0°C or 273K), and T2 is the final temperature of ice (1°C or 274K).
We also know that the enthalpy of vaporization of water is 80 cal/gm.
Now, we need to find the change in pressure (P2 - P1) to change the melting point of ice by 1K. Let's assume P1 is the atmospheric pressure (1 atm or 760 mmHg).
ln(P2/1 atm) = (80 cal/gm)/(R) * (1/273K - 1/274K)
To convert the specific volume from cal/gm to J/kg, we multiply by 4184 J/cal.
ln(P2/1 atm) = (80 cal/gm * 4184 J/cal)/(R * 273K * 1000 gm/kg) * (1/273K - 1/274K)
Simplifying the equation:
ln(P2/1 atm) = (80 * 4184)/(273 * R * 1000) * (274 - 273)/(273 * 274)
Now, we can solve for P2:
P2/1 atm = e^((80 * 4184)/(273 * R * 1000) * (274 - 273)/(273 * 274))
Using the ideal gas constant R = 8.314 J/mol·K:
P2/1 atm = e^((80 * 4184)/(273 * 8.314 * 1000) * (274 - 273)/(273 * 274))
P2 ≈ 1.0003 atm
Therefore, to change the melting point of ice by 1K, the change in pressure should be approximately 0.0003 atm.
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To change the melting point of ice by 1K, what should be the change in the pressure, given that L = 80 cal gm-1 and specific volume of ice and water at 0°C are 1 cm3 and 1.091 cm3 respectively. Give your answer in atm.Correct answer is '135.25'. Can you explain this answer?
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To change the melting point of ice by 1K, what should be the change in the pressure, given that L = 80 cal gm-1 and specific volume of ice and water at 0°C are 1 cm3 and 1.091 cm3 respectively. Give your answer in atm.Correct answer is '135.25'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about To change the melting point of ice by 1K, what should be the change in the pressure, given that L = 80 cal gm-1 and specific volume of ice and water at 0°C are 1 cm3 and 1.091 cm3 respectively. Give your answer in atm.Correct answer is '135.25'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for To change the melting point of ice by 1K, what should be the change in the pressure, given that L = 80 cal gm-1 and specific volume of ice and water at 0°C are 1 cm3 and 1.091 cm3 respectively. Give your answer in atm.Correct answer is '135.25'. Can you explain this answer?.
To change the melting point of ice by 1K, what should be the change in the pressure, given that L = 80 cal gm-1 and specific volume of ice and water at 0°C are 1 cm3 and 1.091 cm3 respectively. Give your answer in atm.Correct answer is '135.25'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about To change the melting point of ice by 1K, what should be the change in the pressure, given that L = 80 cal gm-1 and specific volume of ice and water at 0°C are 1 cm3 and 1.091 cm3 respectively. Give your answer in atm.Correct answer is '135.25'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for To change the melting point of ice by 1K, what should be the change in the pressure, given that L = 80 cal gm-1 and specific volume of ice and water at 0°C are 1 cm3 and 1.091 cm3 respectively. Give your answer in atm.Correct answer is '135.25'. Can you explain this answer?.
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