A container holds 1.3 mol of gas. The total average kinetic energy of the gas molecules in the container is equal to the kinetic energy of a 6.7x10-3 kg bullet with a speed of 830 m/s. What is the Kelvin temperature of the gas?
To find the Kelvin temperature of the gas, we can use the equation for average kinetic energy:
KE = (3/2) * k * T
Where:
KE = average kinetic energy
k = Boltzmann constant (1.38 x 10^-23 J/K)
T = temperature in Kelvin
We can find the average kinetic energy of the gas by equating it to the kinetic energy of the bullet:
KE_gas = KE_bullet
(3/2) * k * T_gas = (1/2) * m_bullet * v_bullet^2
Where:
m_bullet = mass of the bullet
v_bullet = speed of the bullet
Rearranging the equation, we can solve for T_gas:
T_gas = (1/3) * (m_bullet * v_bullet^2) / (k * N)
Where:
N = number of moles of gas
Plugging in the given values:
m_bullet = 6.7 x 10^-3 kg
v_bullet = 830 m/s
k = 1.38 x 10^-23 J/K
N = 1.3 mol
T_gas = (1/3) * (6.7 x 10^-3 kg * 830^2 m^2/s^2) / (1.38 x 10^-23 J/K * 1.3 mol)
Now, let's calculate the temperature.
To find the Kelvin temperature of the gas, we can use the equation:
KE = (3/2) * n * R * T
Where:
KE = Total average kinetic energy of the gas molecules
n = Number of moles of the gas
R = Gas constant (8.314 J/(mol·K))
T = Kelvin temperature
First, let's calculate the energy of the bullet:
KE_bullet = (1/2) * m * v^2
Where:
KE_bullet = Kinetic energy of the bullet
m = Mass of the bullet (6.7x10^-3 kg)
v = Speed of the bullet (830 m/s)
KE_bullet = (1/2) * (6.7x10^-3 kg) * (830 m/s)^2
Now, let's calculate the kinetic energy of the gas molecules:
KE_gas = KE_bullet
Since the total average kinetic energy of the gas molecules is equal to the kinetic energy of the bullet.
Next, let's rearrange the equation to solve for the temperature (T):
T = (KE_gas) / ((3/2) * n * R)
Now we can substitute the values into the equation and solve for T:
T = ((1/2) * (6.7x10^-3 kg) * (830 m/s)^2) / ((3/2) * (1.3 mol) * (8.314 J/(mol·K)))
Calculating this expression will give us the Kelvin temperature of the gas.