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? ( PV = nRT; R= 8.31 L-kPa/mol-K)
Substitute and solve .
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 of gas
R = ideal gas constant (8.31 L-kPa/mol-K)
T = temperature (in Kelvin)
Given:
P = 100.0 kPa
n = 0.050 mol
R = 8.31 L-kPa/mol-K
T = 295 K
Now, we can substitute the given values into the equation and solve for V:
V = (nRT) / P
V = (0.050 mol * 8.31 L-kPa/mol-K * 295 K) / 100.0 kPa
V = 1.21755 L
Therefore, the gas will occupy a volume of 1.21755 liters under the given conditions.
To find the volume of the gas under the given conditions, we can use the ideal gas law equation:
PV = nRT
Where:
P is the pressure of the gas (in kPa)
V is the volume of the gas (in liters)
n is the number of moles of gas
R is the ideal gas constant (8.31 L-kPa/mol-K)
T is the temperature of the gas (in Kelvin)
Let's plug in the values given in your question:
P = 100.0 kPa
n = 0.050 moles
R = 8.31 L-kPa/mol-K
T = 295 K
Now, rearranging the equation to solve for V:
V = (nRT) / P
Substituting the given values into the equation:
V = (0.050 mol * 8.31 L-kPa/mol-K * 295 K) / 100.0 kPa
Calculating:
V = (12.2535 L-kPa) / 100.0 kPa
V ≈ 0.1225 L
Therefore, the gas will occupy approximately 0.1225 liters under these conditions.