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A volatile liquid with vapor pressure 85 kPa (at sea level, 25° C) is taken to the peak of Mt. Everest. Which of the following is true?
- a)The vapor pressure of the solution decreases
- b)The solution will condense quickly than when at sea level
- c)The solution will vaporize quickly than when at sea level
- d)The vapor pressure of the solution increases
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
A volatile liquid with vapor pressure 85 kPa (at sea level, 25° C)...
°C) is placed in a closed container. The container is then heated to a temperature of 100°C. What will be the vapor pressure of the liquid at this new temperature?
To determine the vapor pressure of the liquid at the new temperature, we can use the Clausius-Clapeyron equation:
ln(P2/P1) = (ΔHvap/R) * (1/T1 - 1/T2)
Where:
P1 = initial vapor pressure (85 kPa)
P2 = final vapor pressure (unknown)
ΔHvap = enthalpy of vaporization
R = ideal gas constant (8.314 J/mol·K)
T1 = initial temperature (25+273 = 298 K)
T2 = final temperature (100+273 = 373 K)
First, let's calculate the value of ΔHvap for the liquid. This information is usually provided or can be looked up in a reference source. For this calculation, let's assume ΔHvap = 40 kJ/mol.
Substituting the known values into the equation:
ln(P2/85) = (40,000 J/mol / 8.314 J/mol·K) * (1/298 K - 1/373 K)
Simplifying:
ln(P2/85) = 4801.2 * (0.0034 - 0.0027)
ln(P2/85) = 4801.2 * 0.0007
ln(P2/85) = 3.36084
Using the property of logarithms, we can rewrite the equation as:
P2/85 = e^3.36084
Solving for P2:
P2 = 85 * e^3.36084
Using a scientific calculator or an online calculator, we find:
P2 ≈ 412.57 kPa
Therefore, the vapor pressure of the liquid at the new temperature of 100°C is approximately 412.57 kPa.
To determine the vapor pressure of the liquid at the new temperature, we can use the Clausius-Clapeyron equation:
ln(P2/P1) = (ΔHvap/R) * (1/T1 - 1/T2)
Where:
P1 = initial vapor pressure (85 kPa)
P2 = final vapor pressure (unknown)
ΔHvap = enthalpy of vaporization
R = ideal gas constant (8.314 J/mol·K)
T1 = initial temperature (25+273 = 298 K)
T2 = final temperature (100+273 = 373 K)
First, let's calculate the value of ΔHvap for the liquid. This information is usually provided or can be looked up in a reference source. For this calculation, let's assume ΔHvap = 40 kJ/mol.
Substituting the known values into the equation:
ln(P2/85) = (40,000 J/mol / 8.314 J/mol·K) * (1/298 K - 1/373 K)
Simplifying:
ln(P2/85) = 4801.2 * (0.0034 - 0.0027)
ln(P2/85) = 4801.2 * 0.0007
ln(P2/85) = 3.36084
Using the property of logarithms, we can rewrite the equation as:
P2/85 = e^3.36084
Solving for P2:
P2 = 85 * e^3.36084
Using a scientific calculator or an online calculator, we find:
P2 ≈ 412.57 kPa
Therefore, the vapor pressure of the liquid at the new temperature of 100°C is approximately 412.57 kPa.
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Community Answer
A volatile liquid with vapor pressure 85 kPa (at sea level, 25° C)...
The vapor pressure of a liquid is independent of the pressure of surroundings. Any solution boils when its vapor pressure equals the surrounding atmospheric pressure. The atmospheric pressure decreases with increase in altitude. Due to this, the vapor pressure equals the atmospheric pressure in a shorter period of time when compared to at the solution being present at sea level. Therefore, the solution will vaporize quickly as the bowling point will reached faster.
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A volatile liquid with vapor pressure 85 kPa (at sea level, 25° C) is taken to the peak of Mt. Everest. Which of the following is true?a)The vapor pressure of the solution decreasesb)The solution will condense quickly than when at sea levelc)The solution will vaporize quickly than when at sea leveld)The vapor pressure of the solution increasesCorrect answer is option 'D'. Can you explain this answer?
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