If 4.65 L of CO2 gas at 23 C at 779 mmHg is used, what is the final volume, in l
ID: 1006364 • Letter: I
Question
If 4.65 L of CO2 gas at 23 C at 779 mmHg is used, what is the final volume, in liters, of the gas at 35 C and a pressure of 753 mmHg , if the amount of CO2 remains the same? /^^^^^^^^^^/ What is the volume, in L, of 36.0 g of CH4 at STP? /^^^^^^^^^/ Use molar volume to calculate each of the following at STP: the number of moles of O2 in 48.8 L of O2 gas & the volume, in liters, occupied by 0.570 mole of He gas? /^^^^^^^^^^^^/ In a gas mixture, the partial pressures are argon 420 mmHg , neon 95 mmHg , and nitrogen 140 mmHg .What is the total pressure (atm) exerted by the gas mixture? /^^^^^^^^^^^^^^^/ A mixture of nitrogen (N2) and helium has a volume of 250 mL at 36 C and a total pressure of 763 mmHg .If the partial pressure of helium is 22 mmHg , what is the partial pressure of the nitrogen? & At this temperature and at the partial pressure in Part A, how many moles of nitrogen gas are in this container assuming ideal conditions? please help!!! need by tomorrow
Explanation / Answer
Applying ideal gas law for all the problems given
1 ) P1V1/T1=P2V2/T2
779*4.65/296 =753*V2/308
V2=5.0006L
2)PV=nRT
AT STP,
1*V=(36/44)*0.0821*273.15
V=18.348L
3)volume occupied by 1 mole of any ideal gas =22.4L at STP
given 48.8 L of oxygen gas has 48.8/22.4 moles of O2 i.e ,
2.178 moles of oxygen
1 mole of He gas occupies 22.4 L vol
0.570 mole of He gas occupies 0.570*22.4 = 12.768 L volume
4)according to daltons law , total pressure of gaseous mixture is equal to sum of partial pressures exerted by individual gases
Therefore total pressure =420+95+140= 655mm of Hg
5) partial pressure of nitrogen gas =total pressure -partial pressure of helium gas
I.e, 763-22 = 741mm of Hg
Partial pressure = molefraction*total pressure
Therefore molefraction of N2 gas = 741/763=0.9711
From PV=nRT,
1.0039 atm *250 ml/1000= n*0.0821*309
n= 9.8930 moles (total)
No of moles of N2 gas = molefraction *total moles
= 0.9711*9.8930=9.607moles