Gases Exam 3 and Problem Solutions

Gases Exam 3 and Problem Solutions

1. Find volume of 0,5mol CH₄under 3,28 atm pressure and 400 ⁰K temperature.

Solution:

P=3,28 atm, n=0,5mol, T=400 ⁰K, R=0,082, V=?

We use ideal gas law;

P.V=n.R.T

3,28.V=0,5.0,082.400

V=5 liters

2. If 6,4 g CH₄ has pressure 0,5 atm and volume 2 liters, find pressure of 9 g C₂H₆ having 1 liter volume under constant temperature.(C=12, H=1)

Solution:

We first find mole of given matters;

n_CH4=6,4/16=0,4mol

n_C2H6=9/30=0,3mol

Since temperature is constant we write ideal gas law as given below;

(0,5.2)/0,4=(P₂.1)/0,3

P₂=0,75 atm

or P₂= 57 cm Hg

3 . Find density of O₂ under 27 ⁰C temperature and 1,23 atm pressure. (O=16)

Solution:

T=27 + 273=300 ⁰K

If we write ideal gas law for density, we get following equation;

d=(P.M)/(R.T)

where M is molar mass of O₂.

d=(1,23.32)/(0,082.300)

d=1,6g/liter

4. If we open the taps given in the picture below, find final temperature of gases.

Solution:

We use following equation to find final pressure of gas mixture;

P₁.V₁ + P₂.V₂ + P₃.V₃ = P_final.V_final

3.2 + 4.3 + 0.5 =P_final.(2+3+5)

6 + 12=P_finall.10

P_final=1,8 atm

5. When we open the taps given in the picture below, find changes in the pressures of gases.

Solution:

We should find final pressure of system to make comparison.

P₁.V₁ + P₂.V₂ + P₃.V₃ = P_final.V_final

P.V + 2P.2V + 3P.V = P_final.(V+2V+V)

P_final=2P

Thus,

I. Pressure of first container increase

II. Pressure of second container stays constant

III. Pressure of third container decreases