A 1.00 L sample of a pure gas weighs 0.785 g and is at 733.4 torr and 29.2 degree celcius.
a) what is the molar mass of the gas?
b) if the volume of the temperature are kept constant while 0.400 g of the same gas are added to that already in the container, what will the new pressure be?
Note the correct spelling of celsius.
1). Use PV = nRT and solve for n = number of moles, then n = grams/molar mass. You have n and grams, solve for molar mass. Don't forget to convert T to kelvin and P to atmospheres.
2) Add 0.400 g to grams already there, use the molar mass from the first part to convert grams to moles, then use PV = nRT and solve for pressure in atmospheres.
To find the molar mass of the gas, we can use the Ideal Gas Law equation:
PV = nRT
Where:
P = pressure in atm
V = volume in liters
n = number of moles of gas
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature in Kelvin
First, let's convert the given conditions to appropriate units:
Pressure: 733.4 torr = 733.4/760 atm (since 1 atm = 760 torr)
Volume: 1.00 L (already in liters)
Temperature: 29.2 degrees Celsius = 29.2 + 273.15 Kelvin (since Kelvin = Celsius + 273.15)
Now, we can plug in the values into the Ideal Gas Law equation to solve for moles (n):
(733.4/760) * 1.00 = n * 0.0821 * (29.2 + 273.15)
Simplifying the equation:
0.963 = n * 22.4145
n = 0.963 / 22.4145
n ≈ 0.043 moles
To find the molar mass (M) of the gas, we can use the formula:
Molar mass = mass (g) / moles (n)
Given mass: 0.785 g
Molar mass = 0.785 g / 0.043 moles
Molar mass ≈ 18.26 g/mol
a) Therefore, the molar mass of the gas is approximately 18.26 g/mol.
Now, moving on to the second question:
b) If 0.400 g of the same gas is added, the moles of the gas will increase. We can find the new moles (n) using the formula:
moles of added gas = mass of added gas / molar mass
Given mass: 0.400 g
Molar mass (calculated in part a): 18.26 g/mol
moles of added gas = 0.400 g / 18.26 g/mol
moles of added gas ≈ 0.022 moles
Since volume and temperature are kept constant, the new total moles of gas in the container will be the sum of the original moles and the moles of added gas:
Total moles = 0.043 moles (original gas) + 0.022 moles (added gas)
Total moles ≈ 0.065 moles
To find the new pressure, we can rearrange the Ideal Gas Law equation:
P1V1 = P2V2
Where:
P1 = initial pressure
V1 = initial volume
P2 = new pressure
V2 = V1 (since volume is constant)
Rearranging the equation:
P2 = (P1 * V1) / V2
Plugging in the values:
P1 = 733.4/760 atm (initial pressure)
V1 = 1.00 L (initial volume)
V2 = 1.00 L (new volume, kept constant)
Total moles = 0.065 moles (calculated earlier)
P2 = [(733.4/760) * 1.00] / 1.00
P2 ≈ 0.965 atm
b) Therefore, the new pressure will be approximately 0.965 atm.