How many moles of gas are in a 6.00 L container at a temperature of 360 K and a pressure of 2.35 atm? Given R= 0.0821 L atm/mol K.
To calculate the number of moles of gas, we can use the ideal gas law equation:
PV = nRT
Where:
P = Pressure (in atm)
V = Volume (in liters)
n = Number of moles
R = Ideal gas constant (0.0821 L atm/mol K)
T = Temperature (in Kelvin)
Rearranging the equation to solve for n, we have:
n = PV / RT
Plugging in the given values:
P = 2.35 atm
V = 6.00 L
R = 0.0821 L atm/mol K
T = 360 K
n = (2.35 atm * 6.00 L) / (0.0821 L atm/mol K * 360 K)
n = (14.1 L atm) / (29.556 L atm/mol K)
n ≈ 0.4767 moles
Therefore, there are approximately 0.4767 moles of gas in the 6.00 L container.
To find the number of moles of gas in a given container, you can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles of gas
R = ideal gas constant (0.0821 L atm/mol K)
T = temperature (in Kelvin)
In this case, you are given:
P = 2.35 atm
V = 6.00 L
T = 360 K
R = 0.0821 L atm/mol K
Substituting the given values into the equation, you can solve for n:
(2.35 atm) × (6.00 L) = n × (0.0821 L atm/mol K) × (360 K)
14.10 L atm = n × 29.556 L atm/mol
Dividing both sides of the equation by 29.556 L atm/mol, we get:
n = 14.10 L atm / 29.556 L atm/mol
n = 0.4776 mol
Therefore, there are approximately 0.48 moles of gas in the 6.00 L container at a temperature of 360 K and a pressure of 2.35 atm.
To find the number of moles of gas, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in L)
n = number of moles
R = gas constant (0.0821 L atm/mol K)
T = temperature (in Kelvin)
Rearranging the equation to solve for n:
n = PV / RT
Substituting the given values:
P = 2.35 atm
V = 6.00 L
R = 0.0821 L atm/mol K
T = 360 K
n = (2.35 atm * 6.00 L) / (0.0821 L atm/mol K * 360 K)
Simplifying the expression:
n = 14.1 / 29.556
n ≈ 0.477 moles
Therefore, there are approximately 0.477 moles of gas in a 6.00 L container at a temperature of 360 K and a pressure of 2.35 atm.