Question

Name:_____________________________________________________ Date:___________ Period:_________ 3/23 - 3/27 Assignment 1: Gas Law Practice Problems 1. A gas occupies 12.3 liters at a pressure of 40.0 mm Hg. What is the volume when the pressure is increased to 60.0 mm Hg? Given: Equation: 2. A gas occupies 900.0 mL at a temperature of 27.0 °C. What is the volume at 132.0 °C? Given: Equation: 3. If a gas in a closed container is pressurized from 15.0 atmospheres to 16.0 atmospheres and its original temperature was 25.0 °C, what would the final temperature of the gas be? Given: Equation: 4. A sample of gas occupies 2.00 l with 5.00 moles present. What would happen to the volume if the number of moles is increased to 10.0? Given: Equation:

Answer 1

Answer:

The answers are;

1. 8.2 liters

2. 1214.84 ml

3. 318.027 K

4. 4.00 l.

Explanation:

1. Boyle's law states that the volume of  given mass of gas is inversely proportional to its pressure at constant temperature

that is

P₁·V₁ = P₂·V₂

Where:

P₁ = Initial pressure = 40.0 mm Hg

V₁ = Initial volume = 12.3 liters

P₂ = Final pressure = 60.0 mm Hg

V₂ = Final volume = Required

From P₁·V₁ = P₂·V₂, V₂ is given by

$V_2=\frac{P_1\cdot V_1}{P_2} = \frac{40.0 mm Hg\cdot 12.3 l}{60.0 mm Hg}$ =  8.2 l

The volume reduces to V₂ = 8.2 liters

2. Here Charles law states that

$\frac{T_1}{V_1} =\frac{T_2}{V_2}$

T₁ = Initial temperature = 27.0 °C = 300.15 K

V₁ = Initial volume = 900.0 mL

T₂ = Final temperature = 132.0 °C = 405.15 K

V₂ = Final volume = Required

Therefore  $V_2 =\frac{T_2\cdot V_1}{T_1}$ = $\frac{405.15 K\times 900.0 mL}{300.15 K}$ = 1214.84 ml

V₂ = 1214.84 ml

3.  Gay-Lussac's Law states that

$\frac{T_1}{P_1} =\frac{T_2}{P_2}$

Where:

P₁ = Initial pressure = 15.0 atmospheres

T₁ = Initial temperature = 25.0 °C = 298.15 K

P₂ = Final pressure = 16.0 atmospheres

T₂ = Final temperature = Required

∴ $T_2 = \frac{T_1\times P_2}{P_1}$

=  $\frac{298.15 K\times 16.0atm}{15.0atm}$ = 318.027 K

T₂ = 318.027 K

4. Avogadro's law states that,

Equal volume of all gases at the same temperature and pressure contain equal number of molecules.

Therefore if 5.00 moles of gas occupies 2.00 l volume, then

1 moles will occupy 2.00/5 l volume and

10 moles will occupy 2.00/5 × 10 or 4.00 l volume.