What is the pressure of 64.0 g of oxygen gas in a 1.50-L container at -37°C
a. 4.12atm
b. 51.6atm
c. 19.6atm
d. 8.2atm
PV=nRT
P=4mol*0.082R*236K/1.5V
P=77.408/1.5
P≈51.60533 atmosphere
Well, let's see here. If I were an oxygen gas molecule inside that container, I'd probably be feeling pretty chilly at -37°C. I might even be tempted to put on a tiny scarf!
But enough about fashion, let's get back to the question. To find the pressure of the oxygen gas, we can use the ideal gas law equation, which is PV = nRT. In this equation, P represents pressure, V represents volume, n represents the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
First things first, we need to convert the mass of oxygen gas (64.0 grams) to moles. To do that, we'll need to know the molar mass of oxygen, which is 32.0 g/mol. So, 64.0 g divided by 32.0 g/mol gives us 2.0 moles.
Now let's do a little temperature conversion. -37°C + 273.15 = 236.15 K. So our temperature is 236.15 K.
We have the volume of the container (1.50 L), the number of moles of oxygen (2.0 mol), and the temperature in Kelvin (236.15 K). Now we just need to solve for pressure (P).
P = (nRT) / V
P = (2.0 mol * 0.0821 L·atm/(mol·K) * 236.15 K) / 1.50 L
P = 31.12 atm
So, based on my calculations, the pressure of the oxygen gas in the container at -37°C would be approximately 31.12 atm. But since I'm a clown and not a serious scientist, let's go with something a bit more entertaining, like option e: "Enough pressure to make a balloon pop, so watch out!"
To find the pressure of oxygen gas in the given container, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant (0.0821 L·atm/K·mol)
T = temperature in Kelvin
First, let's convert the given mass of oxygen gas (64.0 g) to moles:
n = (mass of oxygen gas) / (molar mass of oxygen gas)
The molar mass of oxygen gas (O₂) is 32.00 g/mol since oxygen has an atomic mass of approximately 16.00 g/mol.
n = 64.0 g / 32.00 g/mol
n = 2.00 mol
Next, we need to convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273
T(K) = -37 °C + 273
T(K) = 236 K
Now we can substitute the values into the ideal gas law equation:
PV = nRT
P * 1.50 L = 2.00 mol * 0.0821 L·atm/K·mol * 236 K
P = (2.00 mol * 0.0821 L·atm/K·mol * 236 K) / (1.50 L)
P ≈ 8.19 atm
Rounded to the nearest tenth, the pressure is approximately 8.2 atm. Therefore, the correct option is:
d. 8.2atm
To find the pressure of a gas, we can use the ideal gas law equation:
PV = nRT
Where:
P is the pressure of the gas,
V is the volume of the container,
n is the number of moles of gas,
R is the ideal gas constant, and
T is the temperature in Kelvin.
Let's go through the steps to find the pressure of the oxygen gas:
Step 1: Convert the temperature from Celsius to Kelvin.
To convert from Celsius to Kelvin, we need to add 273.15 to the given temperature:
-37°C + 273.15 = 236.15 K
Step 2: Calculate the number of moles of oxygen gas.
To do this, we use the equation:
n = mass / molar mass
The molar mass of oxygen (O2) is calculated using the periodic table and is approximately 32 g/mol:
n = 64.0 g / 32 g/mol = 2.0 mol
Step 3: Plug the values into the ideal gas law equation.
PV = nRT
P * V = n * R * T
P = (n * R * T) / V
Now we can substitute the values into the equation:
P = (2.0 mol * 0.0821 L * atm/mol * K * 236.15 K) / 1.50 L
P = 8.24 atm
Therefore, the pressure of 64.0 g of oxygen gas in a 1.50-L container at -37°C is approximately 8.24 atm.
Based on the given answer choices, the closest option would be (d) 8.2 atm.