Gases are sold in large cylinders for laboratory use. What pressure in atmospheres will be exerted by 2500 g of oxygen gas (O2) when stored at 22°C in a 40.0 L cylinder?

Question

Wyatt Atkins · Accepted Answer

47.3 atm

Assuming near ideal gas behavior where #n = (PV)/(RT)#, with a constant volume, we use the given values with the appropriate gas constant value R to find the pressure. Rearranging the equation we have #P = (n*R*T)/V # V = 40.0 L R = 0.0820575 L-atm/K-mol T = 22'C = 295'K n = (2500g/32.0 g/mol) = 78.125 moles of #O_2#. #P = (78.125*0.0820575*295)/40.0# P is equal to 47.3 atm.

Kai Bowman · Answer

undefined To find the pressure exerted by the oxygen gas (O2), 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, we need to convert the temperature from Celsius to Kelvin:

22°C + 273.15 = 295.15 K

Next, we calculate the number of moles of oxygen gas using its molar mass:

Molar mass of O2 = 2 * 16.00 g/mol = 32.00 g/mol

Number of moles (n) = mass / molar mass
Number of moles (n) = 2500 g / 32.00 g/mol ≈ 78.125 moles

Now, we plug the values into the ideal gas law equation:

P = nRT / V

P = (78.125 moles * 0.0821 L·atm/mol·K * 295.15 K) / 40.0 L

P ≈ 45.6 atm

So, approximately 45.6 atm of pressure will be exerted by 2500 g of oxygen gas stored at 22°C in a 40.0 L cylinder.