What is the volume of the container needed to store 0.8 moles of argon gas at 5.3 atm and 227°C?
(Given: R = 0.08205 l · atm/mol · K)
6.19 liters.
V=22.4*.8*(273+227)/273 * 1/5.3
= 6.19 liters
To find the volume of the container needed to store 0.8 moles of argon gas at 5.3 atm and 227 °C, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = gas constant (0.08205 l · atm/mol · K)
T = temperature (in Kelvin)
First, convert the temperature from Celsius to Kelvin:
T = 227 °C + 273.15 = 500.15 K
Now we can rearrange the ideal gas law equation to solve for V:
V = (nRT) / P
Substituting the given values:
V = (0.8 moles × 0.08205 l · atm/mol · K × 500.15 K) / 5.3 atm
Simplifying the equation:
V = (33.024 l · atm) / 5.3 atm
V ≈ 6.23 liters
Therefore, the volume of the container needed to store 0.8 moles of argon gas at 5.3 atm and 227°C is approximately 6.23 liters.
To find the volume of the container needed to store a certain amount 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 of gas
R = gas constant (in L·atm/mol·K)
T = temperature (in Kelvin)
First, let's convert the given temperature from Celsius to Kelvin by adding 273.15:
227 °C + 273.15 = 500.15 K
Now, we can plug the values into the ideal gas law equation:
(5.3 atm) * V = (0.8 mol) * (0.08205 L·atm/mol·K) * (500.15 K)
Next, solve for V by dividing both sides of the equation by 5.3 atm:
V = (0.8 mol) * (0.08205 L·atm/mol·K) * (500.15 K) / (5.3 atm)
V ≈ 37.36 liters
Therefore, the volume of the container needed to store 0.8 moles of argon gas at 5.3 atm and 227°C is approximately 37.36 liters.