Question

A bottle contains a mixture of nitrogen and oxygen gases. At a specific temperature and pressure, the nitrogen molecules move on an average velocity of 500 m/s and oxygen molecules move at 400 m/s. What is the partial pressure of nitrogen and oxygen in the mixture if the total pressure of the gas is 1 atm?

Answer 1

To solve this problem, we can use the concept of mole fraction and the kinetic theory of gases. According to the kinetic theory of gases, the average kinetic energy of gas molecules is directly proportional to the temperature. Therefore, we can write: 1/2  * m_N2 * v_N2^2 =  1/2  * m_O2 * v_O2^2where m_N2 and m_O2 are the molar masses of nitrogen and oxygen, respectively, and v_N2 and v_O2 are their average velocities.We know that the molar mass of nitrogen  N2  is 28 g/mol and that of oxygen  O2  is 32 g/mol. The average velocities are given as 500 m/s for nitrogen and 400 m/s for oxygen. Plugging these values into the equation, we get: 1/2  * 28 *  500 ^2 =  1/2  * 32 *  400 ^2Solving for the ratio of the molar masses: 28 *  500 ^2  /  32 *  400 ^2  = n_N2 / n_O2where n_N2 and n_O2 are the moles of nitrogen and oxygen, respectively.Calculating the ratio: 28 *  500 ^2  /  32 *  400 ^2   1.464n_N2 / n_O2  1.464Now, let's find the mole fractions of nitrogen and oxygen:X_N2 = n_N2 /  n_N2 + n_O2 X_O2 = n_O2 /  n_N2 + n_O2 Since n_N2 / n_O2  1.464, we can write n_O2 = 1.464 * n_N2:X_N2 = n_N2 /  n_N2 + 1.464 * n_N2  = 1 /  1 + 1.464   0.406X_O2 = 1 - X_N2  0.594Now, we can find the partial pressures of nitrogen and oxygen using the mole fractions and the total pressure:P_N2 = X_N2 * P_total = 0.406 * 1 atm  0.406 atmP_O2 = X_O2 * P_total = 0.594 * 1 atm  0.594 atmTherefore, the partial pressure of nitrogen in the mixture is approximately 0.406 atm, and the partial pressure of oxygen is approximately 0.594 atm.