A 0.12-mole sample of nitrogen gas occupies a volume of 2.55 L. What is the volume of 0.32 mol of nitrogen gas under the same conditions?
2.55L * .32/.12
To find the volume of 0.32 moles of nitrogen gas under the same conditions, we can use the concept of the ideal gas law.
The ideal gas law equation is:
PV = nRT
Where:
P = pressure
V = volume
n = moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin (assumed to be constant)
We have the following information:
For the 0.12-mole sample:
n₁ = 0.12 mol
V₁ = 2.55 L
We want to find:
V₂ (volume for 0.32 mol)
To solve for V₂, we'll use the formula:
(V₁ / n₁) = (V₂ / n₂)
Plugging in the known values:
(2.55 L / 0.12 mol) = (V₂ / 0.32 mol)
Now we can solve for V₂ by rearranging the equation:
V₂ = (2.55 L / 0.12 mol) * 0.32 mol
V₂ = 6.8 L
So, the volume of 0.32 moles of nitrogen gas under the same conditions is 6.8 L.
To find the volume of 0.32 moles of nitrogen gas under the same conditions, we can use the concept of the ideal gas law. The ideal gas law states that the product of pressure (P), volume (V), and the number of moles (n) is equal to the product of the gas constant (R) and temperature (T).
The formula of the ideal gas law is given by:
PV = nRT
where:
P = pressure
V = volume
n = number of moles
R = gas constant
T = temperature
Since the conditions (pressure, temperature) remain the same in this case, we can set up a ratio to find the volume:
V₁ / n₁ = V₂ / n₂
where:
V₁ = initial volume
n₁ = initial number of moles
V₂ = final volume (to be calculated)
n₂ = final number of moles
Plugging in the given values:
V₁ = 2.55 L
n₁ = 0.12 mol
n₂ = 0.32 mol
We can rearrange the formula to solve for V₂:
V₂ = (V₁ * n₂) / n₁
Substituting the known values:
V₂ = (2.55 L * 0.32 mol) / 0.12 mol
Calculating the result:
V₂ = 6.8 L
Therefore, the volume of 0.32 mol of nitrogen gas under the same conditions is 6.8 liters.