7.7 mol of helium are in a 16 {\rm L} cylinder. The pressure gauge on the cylinder reads 65psi. What are (a) the temperature of the gas in Celsius and (b) the average kinetic energy of a helium atom?
7.7 mol of helium are in a 16 L cylinder. The pressure gauge on the cylinder reads 65psi. What are (a) the temperature of the gas in Celsius and (b) the average kinetic energy of a helium atom?
To solve this problem, we need to use the ideal gas law equation, which states that PV = nRT, where P stands for pressure, V for volume, n for the number of moles, R for the ideal gas constant, and T for temperature.
(a) To find the temperature of the gas in Celsius, we need to convert the given pressure from psi to Pascal (Pa), and the given volume from liters (L) to cubic meters (m^3). We can then rearrange the equation as follows:
PV = nRT
First, let's convert the pressure from psi to Pascal:
1 psi = 6894.76 Pa
So, 65 psi = 65 * 6894.76 Pa ≈ 448217 Pa
Second, let's convert the volume from liters to cubic meters:
1 L = 0.001 m^3
So, 16 L = 16 * 0.001 m^3 = 0.016 m^3
Now, we can plug in the values into the ideal gas law equation to solve for temperature:
T = (PV) / (nR)
T = (448217 Pa * 0.016 m^3) / (7.7 mol * 8.314 J/(mol K))
T ≈ 933.3 K
To convert from Kelvin (K) to Celsius (°C), we subtract 273.15:
Temperature in Celsius ≈ 933.3 K - 273.15 ≈ 660.15 °C
So, the temperature of the gas is approximately 660.15 °C.
(b) To find the average kinetic energy (KE) of a helium atom, we can use the equation:
KE = (3/2) * k * T
Where k is the Boltzmann constant (8.314 J/(mol K)).
Let's plug in the values and calculate:
KE = (3/2) * (8.314 J/(mol K)) * 933.3 K
KE ≈ 11,686 J
So, the average kinetic energy of a helium atom is approximately 11,686 joules.