What is the pressure exerted by a .500 mol sample of nitrogen in a 10.0L container at 20C?
PV = nRT
Don't forget T must be in kelvin.
To solve this problem, we can use the ideal gas law equation, which states:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
First, we need to convert the temperature from Celsius to Kelvin. The equation to convert Celsius to Kelvin is:
T(K) = T(°C) + 273.15
So, T = 20°C + 273.15 = 293.15 K
Next, we plug the values into the ideal gas law equation and solve for P:
P * 10.0L = (0.500 mol) * (0.0821 L/mol·K) * 293.15 K
P * 10.0L = 12.955 L·atm
P = 12.955 L·atm / 10.0L
P ≈ 1.3 atm
Therefore, the pressure exerted by the 0.500 mol sample of nitrogen will be approximately 1.3 atmospheres.
To find the pressure exerted by a sample of gas, you can use the Ideal Gas Law equation: PV = nRT, where P represents the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin.
To solve for pressure, rearrange the equation to isolate P:
P = (nRT) / V
Now let's plug in the values given in the question:
n = 0.500 mol (number of moles of nitrogen)
V = 10.0 L (volume of the container)
T = 20°C (temperature in Celsius)
First, we need to convert the temperature from Celsius to Kelvin by adding 273.15:
T = 20 + 273.15 = 293.15 K
Next, we need the value of the ideal gas constant, R. The commonly used value for R is 0.0821 L · atm/(K · mol).
Now plug in the values into the equation:
P = (0.500 mol * 0.0821 L · atm/(K · mol) * 293.15 K) / 10.0 L
By performing the calculation, we can determine the pressure exerted by the nitrogen sample in the container.