0.050 mole of gas were collected in a lab at a temperature of 295 K and a pressure of 100.0 kPa. What volume will the gas occupy under these conditions?
We can use the ideal gas law to solve for the volume:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (8.31 J/mol·K), and T is the temperature in Kelvin.
Plugging in the given values, we get:
V = (nRT)/P
V = (0.050 mol)(8.31 J/mol·K)(295 K)/(100.0 kPa)
V = 0.996 L
Therefore, the gas will occupy a volume of 0.996 L under these conditions.
To find the volume of the gas, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in kPa)
V = volume (in liters)
n = number of moles
R = ideal gas constant (usually 0.0821 L·atm/mol·K)
T = temperature (in Kelvin)
First, let's convert the pressure to atm as the ideal gas constant is usually given in terms of atm:
100.0 kPa = 100.0 kPa * (1 atm / 101.325 kPa) ≈ 0.9869 atm
Next, let's substitute the known values into the ideal gas law equation:
(0.9869 atm) * V = (0.050 mol) * (0.0821 L·atm/mol·K) * (295 K)
Now we can solve for V by dividing both sides of the equation by 0.9869 atm:
V = (0.050 mol) * (0.0821 L·atm/mol·K) * (295 K) / 0.9869 atm
V ≈ 1.240 L
Therefore, the gas will occupy approximately 1.240 liters under these conditions.