00:01
If the gas behaves as an ideal gas, then we can relate its pressure, volume, temperature, and its amount in most using the ideal gas law, which is of the form p, b equals nrt, where p is the pressure, v is the volume, and is the number of moles, r is the gas consent, and t is the temperature in terms of kelvin.
00:22
For this problem, we are given with three gases, which partial pressures are as follows.
00:27
For first gas which is nitrogen gas partial pressure of which is given to be 226 tor the next gas is oxygen partial pressure is 125 and the last gas is helium partial pressure is given to be a 109 tour so well let's read this one down here so 109 now these gases are present in 1 .0.
01:05
50 -liter sample and 1 .50 liter container at the temperature of 25 degrees celsius we wish to find the mass of the gases present in this in the mixture so grams of each gas so we will start by solving for the number of poles and from the valid from the ideal gas equation we have n equals pv over r t so we have to take note that your p here is in terms of atmosphere if your r is in the unit of is equal to 0 .08 to 1 litre atm per mole kelvin.
01:55
Aside from that, your temperature needs to be in degrees or in kelvin.
02:00
So let's convert 25 degrees celsius into kelvin by adding 273, giving us 298 kelvin.
02:07
Then we'll convert the partial pressures into atm by noting that 760 tor is equal to 1 atm.
02:16
So what we will do is we'll divide the partial pressure by 760 giving us 0 .297 for pn2.
02:28
So this is now in atmosphere.
02:30
Doing the same for all the rest we get 0 .164 for oxygen and for helium that's 0 .143.
02:45
Alright so let's solve for n2.
02:49
Here n is equal to pressure of 0.
02:53
Let's check again 0 .297 times 297 times the volume of 1 .5 divided by the gas constant of 0 .08 to 1 times the temperature of 2098 and this gives us the number of moles of 0 .0182...