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PV/T=Constant

Providing the volume remains constant.

P/T=C

35/313=20/T

Therefore T=178.9Kelvin

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The vapor pressure of dichloromethane at 0 degrees Celsius is 134 mmHg The normal boiling point of dichloromethane is 40 degrees Celsius. Calculate the molar heat of vaporization?

To calculate the molar heat of vaporization, we can use the Clausius-Clapeyron equation: ΔHvap = -R * T * ln(P2/P1) Where: ΔHvap = molar heat of vaporization R = gas constant (8.314 J/mol*K) T = temperature difference in Kelvin (40°C + 273 - 0°C = 313 K) P1 = vapor pressure at lower temperature (134 mmHg) P2 = vapor pressure at higher temperature (760 mmHg, since boiling point is at atmospheric pressure) Plugging in the values, we have: ΔHvap = -8.314 * 313 * ln(760/134) = -8.314 * 313 * ln(5.67) ≈ -37.4 kJ/mol Therefore, the molar heat of vaporization of dichloromethane is approximately 37.4 kJ/mol.


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Related Questions

If someones temperature is 104 degrees Fahrenheit what is his temperature in Kelvin?

313 Kelvin.


The vapor pressure of dichloromethane at 0 degrees Celsius is 134 mmHg The normal boiling point of dichloromethane is 40 degrees Celsius. Calculate the molar heat of vaporization?

To calculate the molar heat of vaporization, we can use the Clausius-Clapeyron equation: ΔHvap = -R * T * ln(P2/P1) Where: ΔHvap = molar heat of vaporization R = gas constant (8.314 J/mol*K) T = temperature difference in Kelvin (40°C + 273 - 0°C = 313 K) P1 = vapor pressure at lower temperature (134 mmHg) P2 = vapor pressure at higher temperature (760 mmHg, since boiling point is at atmospheric pressure) Plugging in the values, we have: ΔHvap = -8.314 * 313 * ln(760/134) = -8.314 * 313 * ln(5.67) ≈ -37.4 kJ/mol Therefore, the molar heat of vaporization of dichloromethane is approximately 37.4 kJ/mol.


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If V1 = V2 => p1 / T1 = p2 / T2 <=> p2 = p1 * T2 / T1 Let's assume that T1 = 20 C Degree = 313 K => p2 = p1 * 343/313 => p2 = 1.10 p1 which means 10% higher pressure.


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0oC is equal to 273K, and the magnitude of the individual degrees is the same, for the Celsius scale is used to define K. So 40oC is (40+273)K. = 313K. [And for the pedant, the triple point of water is 273.16K. But that makes no difference to the above.]


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What is the temperature of 25.0 g of chlorine gas with a volume of 6.0 Liters and a pressure of 1.5 ATM?

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