A gas at constant temperature is enclosed in a container with a movable piston. It occupies a space of 10L at a pressure of 500
1 answer:
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Answer:
3/2a
Explanation:
The complete step by step answer is found in the attachment
Answer:
The temperature at which the liquid vapor pressure will be 0.2 atm = 167.22 °C
Explanation:
Here we make use of the Clausius-Clapeyron equation;
Where:
P₁ = 1 atm =The substance vapor pressure at temperature T₁ = 282°C = 555.15 K
P₂ = 0.2 atm = The substance vapor pressure at temperature T₂
= The heat of vaporization = 28.5 kJ/mol
R = The universal gas constant = 8.314 J/K·mol
Plugging in the above values in the Clausius-Clapeyron equation, we have;
T₂ = 440.37 K
To convert to Celsius degree temperature, we subtract 273.15 as follows
T₂ in °C = 440.37 - 273.15 = 167.22 °C
Therefore, the temperature at which the liquid vapor pressure will be 0.2 atm = 167.22 °C.