What is the volume occupied by 0.252mol of helium gas at 1.35atm and 305K ?
PV = nRT
To find the volume occupied by a gas, we can use the ideal gas law, which is given by the equation:
PV = nRT
Where:
P = pressure of the gas (in atm)
V = volume of the gas (in liters)
n = number of moles of the gas
R = ideal gas constant (0.0821 L*atm/(mol*K))
T = temperature of the gas (in Kelvin)
We can rearrange this equation to solve for the volume (V):
V = (nRT) / P
Given values:
n = 0.252 mol
P = 1.35 atm
T = 305 K
Now we can substitute the given values into the equation to calculate the volume:
V = (0.252 mol * 0.0821 L*atm/(mol*K) * 305 K) / 1.35 atm
First, we can cancel out the units and simplify the expression:
V = (0.252 * 0.0821 * 305) / 1.35 L
Next, we can calculate the numerical value:
V = 5.66 L
Therefore, the volume occupied by 0.252 mol of helium gas at a pressure of 1.35 atm and a temperature of 305 K is 5.66 liters.