You measure two different pressures of the sealed vial containing air and ethanol at two different temperatures; 89027 mmHg at 27C
and 90853 mmHg at 40C . From earlier measurements you found that the pressure of air at 27C is 76005 mmHg .
Calculate the VPethanol at both temperatures.
At 27C : mmHg
At 40C : mmHg
130.22
115.56
If the air pressure is 76005 at 27 C, then vp ethanol at 27 must be 89027-76005 = ?
For vp at 40 C I would correct air pressure at 27 to air pressure at 40 (using v1/T1 = v2/T2), then total P - air pressure at 40 C = vp ethanol at 40C.
To calculate the VPethanol at both temperatures, we can use the ideal gas law equation, which states:
PV = nRT
Where:
P is the pressure
V is the volume
n is the number of moles
R is the ideal gas constant (0.0821 L·atm/(K·mol) )
T is the temperature (in Kelvin)
To find the VPethanol at each temperature, we need to compare the pressure of the air with the total pressure of the air and ethanol mixture.
At 27°C:
Given:
Ptotal = 89027 mmHg
Pair (pressure of air) = 76005 mmHg
Vial contains only air and ethanol, so it can be assumed that the total pressure is due to the sum of the partial pressures of air and ethanol.
Thus,
Pairstotal = Pair (partial pressure of air) + VPethanol
Substituting the values:
89027 mmHg = 76005 mmHg + VPethanol
Now, let's solve for VPethanol.
89027 mmHg - 76005 mmHg = VPethanol
13022 mmHg = VPethanol
VPethanol at 27°C = 13022 mmHg
At 40°C:
Given:
Ptotal = 90853 mmHg
Pair (pressure of air) = 76005 mmHg
Similarly, using the same equation as above:
90853 mmHg = 76005 mmHg + VPethanol
Now, let's solve for VPethanol.
90853 mmHg - 76005 mmHg = VPethanol
14848 mmHg = VPethanol
VPethanol at 40°C = 14848 mmHg
So, the VPethanol at 27°C is 13022 mmHg, and the VPethanol at 40°C is 14848 mmHg.
To calculate the vapor pressure of ethanol at both temperatures, we can use the ideal gas law and subtract the partial pressure of air.
The ideal gas law is given by: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
First, convert the temperatures from Celsius to Kelvin:
27°C + 273 = 300K
40°C + 273 = 313K
Next, we need to determine the partial pressure of air at each temperature. Given that the pressure of air at 27°C is 76005 mmHg, we can use this value to subtract it from the total pressure at each respective temperature to get the vapor pressure of ethanol.
To calculate the vapor pressure at 27°C:
VPethanol (27°C) = Total pressure (27°C) - Partial pressure of air (27°C)
= 89027 mmHg - 76005 mmHg
To calculate the vapor pressure at 40°C:
VPethanol (40°C) = Total pressure (40°C) - Partial pressure of air (27°C)
= 90853 mmHg - 76005 mmHg
Now, let's calculate the vapor pressures:
VPethanol (27°C) = 89027 mmHg - 76005 mmHg
= 13022 mmHg
VPethanol (40°C) = 90853 mmHg - 76005 mmHg
= 14848 mmHg
Therefore, the vapor pressure of ethanol at 27°C is 13022 mmHg, and the vapor pressure of ethanol at 40°C is 14848 mmHg.