The volume of a sample of a gas collected over water at 32 Celsius and 752 torr is 627 ml. What will the volume of the gas be when dried and measured at STP?
(Vapor pressure of water at 32 Celsius = 35.7 torr)
Use PV = nRT
For P use
Ptotal = pH2O + pgas.
To find the volume of the gas when dried and measured at STP (Standard Temperature and Pressure), we need to use the ideal gas law equation: PV = nRT.
Given:
Initial volume of the gas (V1) = 627 ml
Initial temperature (T1) = 32°C
Initial pressure (P1) = 752 torr
Vapor pressure of water at 32°C = 35.7 torr
We can convert the temperature from Celsius to Kelvin by adding 273.15:
T1 = 32°C + 273.15 = 305.15 K
Then, using the ideal gas law equation, we can calculate the moles of the gas (n1):
P1V1 = n1RT1
n1 = (P1V1) / (RT1)
Now, we need the gas constant (R), which has a value of 0.0821 L·atm/(mol·K).
Next, we can calculate the volume of the gas at STP. At STP, the temperature is 273.15 K, and the pressure is 1 atm.
P2 = 1 atm
T2 = 273.15 K
Again, using the ideal gas law equation, we can find the final volume (V2):
P2V2 = n1RT2
V2 = (n1RT2) / P2
To find n1, we can use the equation we derived earlier:
n1 = (P1V1) / (RT1)
Substituting the values, we get:
V2 = [(P1V1) / (RT1)] * (RT2 / P2)
Now, let's plug in the values and calculate:
V2 = [(752 torr * 627 ml) / (0.0821 L·atm/(mol·K) * 305.15 K)] * (0.0821 L·atm/(mol·K) * 273.15 K / 1 atm)
= [(752 torr * 627 ml) / (0.0821 L·atm/(mol·K) * 305.15 K)] * (0.0821 L·atm/(mol·K) * 273.15 K)
After converting units, we can find the answer.
Therefore, to find the volume of the gas when dried and measured at STP, substitute the given values into the equation and solve for V2.