8g of hydrogen and oxygen will have kinetic energy in what ratio at 27 celsius?
If temp is same then K.E will be in same ratio i.e 1:1
1:16
To calculate the ratio of kinetic energy for hydrogen and oxygen at a given temperature, we need to use the equation for kinetic energy:
KE = (1/2)mv^2
where KE is the kinetic energy, m is the mass of the particle, and v is the velocity of the particle.
To get the ratio of their kinetic energies, we can cancel out the mass term since the ratio of the masses of hydrogen and oxygen is already known to be 1:16 (hydrogen being lighter than oxygen).
Now, let's calculate the ratio of kinetic energy using the given information:
1. Calculate the velocity of hydrogen (v_h):
We can use the kinetic theory of gases to relate the velocity of gas particles to the temperature. The average velocity (v_avg) of a gas particle is given by:
v_avg = sqrt(3kT / m)
where k is the Boltzmann constant (1.38 x 10^-23 J/K), T is the temperature in Kelvin, and m is the mass of the particle.
Given:
- Temperature (T) = 27 degrees Celsius = 300 Kelvin
- Mass of hydrogen (m_h) = 1 gram = 0.001 kg
Let's plug the values into the equation:
v_h = sqrt(3 * 1.38 x 10^-23 * 300 / 0.001) = 1.657 x 10^4 m/s
2. Calculate the velocity of oxygen (v_o):
Given:
- Mass of oxygen (m_o) = 16 grams = 0.016 kg
Let's calculate:
v_o = sqrt(3 * 1.38 x 10^-23 * 300 / 0.016) = 1.220 x 10^4 m/s
3. Calculate the ratio of kinetic energy (KE_h / KE_o):
Using the equation for kinetic energy, the ratio can be calculated as follows:
(KE_h / KE_o) = (v_h^2 / v_o^2)
Plugging in the calculated values:
(KE_h / KE_o) = (1.657 x 10^4)^2 / (1.220 x 10^4)^2 ≈ 1.92
Therefore, the ratio of kinetic energy for hydrogen and oxygen at 27 degrees Celsius is approximately 1.92.