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Boyles law Charles law Gay-Lussac law Avogadro law Idea Gas law Grahams law Daltons law Rubric: - 35 points - Definition and equation for each of the above gas laws (5x7 = 35 points) 35 points - Solve an example problem & 1 point for solved example problem (5x7 = 35 points). 5 points - I. calculate moles from a given mass (solve one example problem) 5 points - II. calculate mass from given moles (solve one example problem) 10 points - MOLAR VOLUME at STP (solve one example using chemical equations)

Answer 1

Boyles Law :

The absolute pressure exerted by a given mass of an ideal gas is inversely proportional to the volume it occupies if the temperature and amount of gas remain unchanged within a closed system.

Mathematically, Boyle's law can be stated as:

$P\propto {\frac {1}{V}}$

PV = Constant = k

where P is the pressure of the gas, V is the volume of the gas, and k is a constant.

Charles Law :

Charles's law (also known as the law of volumes) is an experimental gas law that describes how gases tend to expand when heated. A modern statement of Charles's law is:

When the pressure on a sample of a dry gas is held constant, the Kelvin temperature and the volume will be in direct proportion.

This relationship of direct proportion can be written as:

.${\displaystyle V\propto T}$

V is the volume of the gas,

T is the temperature of the gas (measured in kelvins).

Gay- Lussac Law :

Gay-Lussac's law states that the pressure of a given mass of gas varies directly with the absolute temperature of the gas, when the volume is kept constant.

Mathematically, it can be written as:

${\displaystyle {\frac {P}{T}}=k}$

P = Pressure of the gas

T= temperature of the gas in kelvin

k= constant

Avogardro Law :

Avogadro's law states that "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules."

Ideal Gas Law :

${\displaystyle PV=nRT,}$

P = Pressure of the gas

T= temperature of the gas in kelvin

V = volume of the gas

n= number of moles of the gas

R = universal gas constant

Graham's Law :

Graham's law of effusion (also called Graham's law of diffusion) was formulated by Scottish physical chemist Thomas Graham in 1848. Graham found experimentally that the rate of effusion of a gas is inversely proportional to the square root of the mass of its particles.This formula can be written as:

${{\mbox{Rate}}_{1} \over {\mbox{Rate}}_{2}}={\sqrt {M_{2} \over M_{1}}}$

where:

Rate1 is the rate of effusion for the first gas. (volume or number of moles per unit time).

Rate2 is the rate of effusion for the second gas.

M1 is the molar mass of gas 1

M2 is the molar mass of gas 2.

Dalton's Law :

Dalton's law (also called Dalton's law of partial pressures) states that in a mixture of non-reacting gases, the total pressure exerted is equal to the sum of the partial pressures of the individual gases.

${\displaystyle p_{\text{total}}=p_{1}+p_{2}+p_{3}+\cdots +p_{n}}$

where p1, p2, ..., pn represent the partial pressures of each component.

Problem 1: A container holds 500 mL of CO2 at 20° C and 742 torr. What will be the volume of the CO2 if the pressure is increased to 795 torr?

solution:P1 = 742 torr

P2 = 795 torr

V1 = 500 mL

V2 = ?

  P1V1 = P2V 2

V2 = P1V1/P2

V2 = 742 torr x 500. mL/795 torr = 467 mL CO2

Problem 2:A container holds 50.0 mL of nitrogen at 25° C and a pressure of 736 mm Hg. What will be its volume if the temperature increases by 35° C?

solution: P1 = 736 mm Hg P2 = 736 mm Hg V1 = 50.0 mL V2 = ? T1 = 25° C + 273 = 298 K T2 = 25° C + 35° C + 273 = 333 K

V1/ T1 = V2/T2

V2 = V1 x T2/T1

V2 = 50.0 mL x 333 K/298 K = 55.9 mL N2

Problem 3:What is the volume of a quantity of a gas at 27oC if its volume was 400mL at 0oC? The pressure remains constant?

solution:

Since the temperature increased, the volume increased. Here we must use a temperature factor greater than 1, that is 300/273

The temperature must be in degrees kelvin.

20oC + 273 = 300 K

0oC + 273 = 273 K

400mL × 300K/273K = 438mL

Problem 4:If 0.950 mol of O2 has a volume of 20.0 L at some temperature and pressure, what would be the volume of 1.350 mol of O2?

solution:n1 = 0.950 mol V1 = 20.0 L

V2 = ?   n2 = 1.350 mol

V1 n1 = V2 n2

V2 = V1n2 / n1 = (20.0 L)(1.350 mol) / 0.950 mol V2 = 28.4 L

Problem 5:What is the volume of a 1 mol of a gas at STP?

solution: V = ? n = 1 mol

T = 273.15 K P = 1 atm

PV = nRT

V= nRT / P = (1 mol)(0.08206 L·atm mol·K)(273.15 K) /1 atm = 22.4 L

Problem 6:Gas X has a molar mass of 72 g/mol and Gas Y has a molar mass of 2 g/mol. How much faster or slower does Gas Y effuse from a small opening than Gas X at the same temperature?

solution:

Graham's Law can be expressed as:

rX(MMX)1/2 = rY(MMY)1/2

where
rX = rate of effusion/diffusion of Gas X
MMX = molar mass of Gas X
rY = rate of effusion/diffusion of Gas Y
MMY = molar mass of Gas Y

We want to know how much faster or slower Gas Y effuses compared to Gas X. To get this value, we need the ratio of the rates of Gas Y to Gas X. Solve the equation for rY/rX.

rY/rX = (MMX)1/2/(MMY)1/2

rY/rX = [(MMX)/(MMY)]1/2

Use the given values for molar masses and plug them into the equation:

rY/rX = [(72 g/mol)/(2)]1/2
rY/rX = [36]1/2
rY/rX = 6

Problem 7:A container contains three gases, N2, O2 and Ar, with partial pressures of 23.3 kPa, 40.9 kPa and 13.7 kPa respectively. What is the total pressure inside the container?

solution:P(N2) = 23.3 kPa P(O2) = 40.9 kPa P(Ar) = 13.7 kPa   P(T )= ?

P(T) = P(N2) + P(O2) + P(Ar)

P(T )= 23.3 kPa + 40.9 kPa + 13.7 kPa = 77.9 kPa

Problem 8: What is the number of moles of 16 g O2

solution: MW of O2 = 32 g

number of moles =16/32 = 0.5 mole

Problem 9: What is the molar mass of  1 mole N2 ?

solution: MW of N2 = 28 g

Molar mass of N2 = 28$\times$1 =28

Problem 10: What will be the volume of carbon dioxide at STP from 2 kg CaCO3?

solution: CaCO3(s) $\rightarrow$ CaO (s) + CO2(g)

the volume of CO2 be = (22.4$\times$ 2000 )/ 100 = 448 L