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Exams and Problem Solutions
--Matters and Properties of Matters Exams and Problem Solutions
--Atomic Structure Exams and Problem Solutions
--Periodic Table Exams and Problem Solutions
--The Mole Concept Exams and Problem Solutions
--Gases Exams and Problem Solutions
--Gases Exam1 and Problem Solutions
--Gases Exam2 and Problem Solutions
--Gases Exam3 and Problem Solutions
--Gases Exam4 and Problem Solutions
--Chemical Reactions Exams and Problem Solutions
--Nuclear Chemistry (Radioactivity) Exams and Problem Solutions
--Solutions Exams and Problem Solutions
--Acids and Bases Exams and Problem Solutions
--Thermochemistry Exams and Problem Solutions
--Rates of Reaction Exams and Problem Solutions
--Chemical Equilibrium Exams and Problem Solutions
--Chemical Bonds Exams and Problem Solutions
Old Version


Gases Exam2 and Problem Solutions

Gases Exam2 and  Problem Solutions

1. Find value of 570 mm Hg in terms of atm.


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.


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.


Pi=380 mm Hg

Vi=1lt=1000 cm3

Vf=200 cm3


We use boyle's law;



Pf=1900 mm Hg=2,5 atm

4. Find pressure of gas under 546 0C, that has pressure 200 mm Hg under 273 0C.


Pi=200 mmHg, Pf=?, Ti=273 0C, Tf=546 0C

We use Guy Lussac Law;


But, we should first convert temperatures from 0C to 0K.

Ti=273 + 273 = 546 0K

Tf= 546 + 273 = 819 0K


Pf=300 mmHg

5. Find pressure of CO2 having 8,8 g mass and 1230 cm3 volume under 27 0C temperature. (CO2=44)


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=4 atm

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Tags: Gas Laws with Examples |  Gases Exam1 and Problem Solutions |  Gases Exam3 and Problem Solutions | 

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