If measurements of a gas are 100L and 300 kilopascals and then the gas is measured a second time and found to be 75L, describe w
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Given that:
- The heat of vaporization = 22.44 kJ/mol = 22440 J/mol
- normal boiling point which is the initial temperature = 0.4° C = (273 + (-0.4))K = 272.6 K
- volume = 250 mL = 0.250 L
- Mass of butane = 0.8 g
- the final temperature = -22° C = (273 + (-22)) K = 251 K
The first step is to determine the vapor pressure at the final temperature of 251K by using the Clausius-Clapeyron equation. This is following by using the ideal gas equation to determine the numbers of moles of butane gas. After that, the mass of butane present in the liquid is determined by using the relation for the number of moles.
Using Clausius-Clapeyron Equation:
where;
P1 and P2 correspond to the temperature at T1 and T2.
∴
replacing the values into the given equation, we have;
As such, at -22° C; the vapor pressure = 0.427 atm
Now, using the ideal gas equation:
PV = nRT
where:
- P = Pressure
- V = volume
- n = number of moles of butane
- R = universal gas constant
- T = temperature
∴
Making (n) the subject of the formula:
We all know that the standard molecular weight of butane = 58.12 g/mol
∴
Using the relation for the number of moles which is:
mass = 0.00518 mole × 58.12 g/mol
mass = 0.301 g
∴
The mass of butane in the flask = 0.301 g
But the mass of the butane present as a liquid in the flask is
= 0.8 g - 0.301 g
= 0.499 g
In conclusion, the mass of the butane present as a liquid in the flask is 0.499 g
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