Gases Exam 2 and Problem Solutions
Gases Exam 2 and Problem Solutions
1. Find value of 570 mm Hg in terms of atm.
Solution:
We know that there is a relation between atm and cm Hg;
1 atm = 76 cm Hg =760 mm Hg
760 mm Hg is 1 atm
570 mm Hg is ? atm
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
?=0,75 atm
2. Atmospheric pressure is 1 atm. Find pressure of gas in terms of atm;
I. If system is closed manometer
II. If system is open manometer.
Solution:
I. If system is closed;
Pgas=210-20=190 mm Hg
760 mm Hg is 1 atm
190 mm Hg is ? atm
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
?=0,25 atm
II. If system is opened;
Pgas=Pair + (210-20)mm Hg
Pair=1 atm=760 mm Hg
Pgas=760 +190 = 950 mm Hg =95 cm Hg
3. Pressure of gas having 1 liter volume is 380 mm Hg. If volume is decreased to 200 cm3, find change in the pressure under constant temperature.
Solution:
Pi=380 mm Hg
Vi=1lt=1000 cm3
Vf=200 cm3
Pf=?
We use boyle’s law;
Pi.Vi=Pf.Vf
380.1000=Pf.200
Pf=1900 mm Hg=2,5 atm
4. Find pressure of gas under 546 0C, that has pressure 200 mm Hg under 273 0C.
Solution:
Pi=200 mmHg, Pf=?, Ti=273 0C, Tf=546 0C
We use Guy Lussac Law;
Pi/Ti=Pf/Tf
But, we should first convert temperatures from 0C to 0K.
Ti=273 + 273 = 546 0K
Tf= 546 + 273 = 819 0K
200/546=Pf/819
Pf=300 mmHg
5. Find pressure of CO2 having 8,8 g mass and 1230 cm3 volume under 27 0C temperature. (CO2=44)
Solution:
mole of CO2 =mass/molar mass=8,8/44=0,2mol
T=27 + 273=300 0K
V=1230 cm3 =1,23 liter
Ideal gas law is used;
P.V=n.R.T
P.1,23=0,2.0,082.300
P=4 atm