Neon gas produced for use in luminous tubes comes packaged in 1.0-L containers at a pressure of 1.02 atm and a temperature of 24°C. How many moles of Ne do these cylinders hold?
Use PV = nRT and solve for n = number of mols.;
A gas can holds 4.3 gal of gasoline. What is this quantity in cubic centimeters
To calculate the number of moles of neon gas in the cylinder, we can use the ideal gas equation, which is:
PV = nRT
where:
P is the pressure (in atm)
V is the volume (in liters)
n is the number of moles
R is the ideal gas constant (0.0821 L·atm/(mol·K))
T is the temperature (in Kelvin)
First, we need to convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 24 °C + 273.15 = 297.15 K
Now, let's plug in the values into the ideal gas equation:
(1.02 atm)(1.0 L) = n(0.0821 L·atm/(mol·K))(297.15 K)
Simplifying:
1.02 = n(0.0821)(297.15)
Dividing both sides by (0.0821)(297.15):
n = (1.02 atm)/(0.0821 L·atm/(mol·K) · 297.15 K)
Calculating:
n ≈ 0.041 moles
Therefore, the cylinder contains approximately 0.041 moles of neon gas.
To find the number of moles of neon gas in the cylinders, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atmospheres, atm)
V = volume (in liters, L)
n = number of moles of gas
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin, K)
First, we need to convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 24°C + 273.15 K = 297.15 K
Now, we can substitute the given values into the ideal gas law equation:
(1.02 atm)(1.0 L) = n(0.0821 L·atm/(mol·K))(297.15 K)
Simplifying the equation, we get:
1.02 = n(0.0821)(297.15)
1.02 = n(24.455715)
n = 1.02/24.455715
n ≈ 0.0417 mol
So, the cylinders hold approximately 0.0417 moles of neon gas.