A gas mixture contains each of the following gases at the indicated partial pressures: N2, 215 torr; O2, 102 torr; and He, 117 torr. What is the total pressure of the mixture? What mass of each gas is present in a 1.35-L sample of this mixture at 25.0 °C?

Video transcript

Hello everyone today with the falling problem. A gas mixture contains each of the falling gasses at the indicated partial pressures. What is the total pressure of the mixture? What mass of each gas is present in a 1.75 L sample of this mixture at 25 °C. So first, I want to just convert the degrees from Celsius to degrees Kelvin. Such that we have 2 98.15 degrees Kelvin. So now we can solve for the total pressure which will be equal to the partial pressures of each gas, which is the partial pressure of the nitrogen gas plus the partial pressure of the oxygen gas plus the partial pressure of our helium. So for our nitrogen gas, we have 235 tor, we add that to 114 tor for the oxygen gas. And then lastly, we add our 138 torque from the helium gas. This gives us a total pressure of 487 tour. So we can eliminate, enter twice C. Now we use the ideal gas equation to find the mass. And so how does that look? Well, we have the pressure multiplied by the volume is equal to the number of moles multiplied by a gas content are multiplied by the temperature. In Kelvin, the number of moles can be said to be equivalent to the mass divided by the molar mass. So if we rearrange that equation to solve it for M or the, in the little mass, we have M is equal to N multiplied by M. So going back to our ideal gas law, we have M is equal to our more mass multiplied by our pressure multiplied by the volume divided by our gas constant and multiplied by our temperature. So now we can find the individual masses. So first, we will work on our nitrogen gas. Applying that equation, we have the M molar mass, which if we take two individual elements of nitrogen on the pair table, we get 28.02 g per mole. We multiply this by its pressure which was 235 tour, we convert that tor into atmospheres by using the conversion factor that one atmosphere is divided by 760 tor. And let me multiply that by our volume which is 1.75 L, we then multiply by a gas consonant which can be found in the reference text as 0.0 82 06 Lear times atmospheres divided by walls, times Kelvin. And then lastly multiplied by the temperature temperature we sold earlier was 2 98.15 at Kelvin. When all of our units canceled out. We get an individual mass of point 620 grahams. So we can eliminate entero a moving on to our second guess we can do the mass for the oxygen. And we use the same formula if we take two individual elements of oxygen from the par table and we add their molar masses, we get 32 g per mole multiply that by the oxygen gasses pressure, which was 114 tour. And then we convert that to a series by multiplying by the conversion factor. Then lastly by the 1.75 L. And then we divide that by the same values that we had in our first equation for solving for the mass. And so when we plug these numbers in and calculate for the mass of oxygen gas, we get 0.3 43 g. And last but not least we have the mass for our helium. So we do the same procedure where we have our mass of 4.003 g, promo multiplying it by the partial pressure, which is 138 tor. And then we multiply that by the conversion factor, that one atmosphere is divided by 760 tor and then 1.75 L, we divide that by the same values 0.08206 L times atmospheres divided by moles times Kelvin multiplied by 2 98.15 degrees. Kelvin for a mass that is equal to 0.0520 g. And if you look at our choices between Ritos V and D, we see that d best reflects these values. And with that, we have solved the problem. Overall, I hope is helped. And until next time.