# Gases Exam 3 and Problem Solutions

**Gases Exam 3 and Problem Solutions**

**1.** Find volume of 0,5mol CH_{4}under 3,28 atm pressure and 400 ^{0}K temperature.

Solution:

P=3,28 atm, n=0,5mol, T=400 ^{0}K, 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 CH_{4} has pressure 0,5 atm and volume 2 liters, find pressure of 9 g C_{2}H_{6} 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=(P_{2}.1)/0,3

P_{2}=0,75 atm

or P_{2}= 57 cm Hg

**3** . Find density of O_{2} under 27 ^{0}C temperature and 1,23 atm pressure. (O=16)

**Solution:**

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

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

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

where M is molar mass of O_{2}.

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;

P_{1}.V_{1} + P_{2}.V_{2} + P_{3}.V_{3} = P_{final}.V_{final}

3.2 + 4.3 + 0.5 =P_{final}.(2+3+5)

6 + 12=P_{final}l.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.

P_{1}.V_{1} + P_{2}.V_{2} + P_{3}.V_{3} = 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