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## Gases Exam2 and Problem Solutions

**Gases Exam2 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 cm^{3}, find change in the pressure under constant temperature.

**Solution:**

Pi=380 mm Hg

Vi=1lt=1000 cm^{3}

Vf=200 cm^{3}

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 ^{0}C, that has pressure 200 mm Hg under 273 ^{0}C.

**Solution:**

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

We use Guy Lussac Law;

Pi/Ti=Pf/Tf

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

Ti=273 + 273 = 546 ^{0}K

Tf= 546 + 273 = 819 ^{0}K

200/546=Pf/819

Pf=300 mmHg

**5. **Find pressure of CO_{2} having 8,8 g mass and 1230 cm^{3} volume under 27 ^{0}C temperature. (CO_{2}=44)

**Solution:**

mole of CO_{2} =mass/molar mass=8,8/44=0,2mol

T=27 + 273=300 ^{0}K

V=1230 cm^{3} =1,23 liter

Ideal gas law is used;

P.V=n.R.T

P.1,23=0,2.0,082.300

P=4 atm

The Original Author: Mrs. Şerife (Erden) SARICA

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

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