Chapter 11 - Gases

¡Supera tus tareas y exámenes ahora con Quizwiz!

Given: T1 = 298K; V1 = 2.37L; T2 = 354K Find: V2

(2.37L)/298K = V2/354K .007953 L/K = V2/354K V2 = 2.82 L

Convert each measurement to atm. a) 1277 mm Hg b) 2.38 X 10^5 Pa c) 127 psi d) 455 torr

1 atm = 760 mm Hg 1 atm = 101,325 Pa 1 atm = 14.7 psi 1 atm = 760 torr a) 1277 mm Hg 1277 mm Hg X 1 atm/760 mm Hg = 1.68 atm b) 2.38 X 10^5 Pa 2.38 X 10^5 Pa X 1 atm/101325 Pa = 2.35 atm c) 127 psi 127 psi X 1 atm/14.7 psi = 8.64 atm d) 455 torr 455 torr X 1 atm/760 torr = 0.599 atm

Calculate the number of liters of CO2 gas that forms at STP when 0.879 moles of CaCO3 undergoes this reaction: CaCO3 (s) ---> CaO (s) + CO2 (g) Given: 0.879 mole CaCO3 Find: CO2 (g) in liters

1 mol CaCO3 = 1 mol CO2 1 mol CO2 = 22.4 L CO2 0.879 mol CaCO3 X 1 mol CO2/1 mol CaCO3 = 0.879 mol CO2 0.879 mol CO2 X 22.4 L/1 mol CO2 = 19.6896 L CO2 = 19.7 L CO2

If each gas sample has the same temperature and pressure, which has the greatest volume? a) 1 g O2 b) 1 g Ar c) 1 g H2

1 mol of O2 = 31.998 g 1 mol of Ar = 39.948 g 1 mol of H2 = 2.0158 g Since hydrogen gas has the lowest molar mass of the set, 1 g will have the greatest number of moles and therefore the greatest volume. C) 1 g H2

What is Dalton's Law of Partial Pressure?

A law stating that the sum of the partial pressures of each component in a gas mixture equals to total pressure. P total= Pa + Pb + Pc ....

What is Avogadro's Law?

A law stating that the volume (V) of a gas and the amount of the gas in moles (n) are directly proportional. V1/n1 = V2/n2 We can can use Avogadro's Law to calculate the volume of a gas following a change the the amount of the gas as long as the pressure and temperature of the gas are constant.

A flask contains a gas sample at pressure x. If the volume of the container triples at a constant temperature and constant amount of gas, the pressure becomes: a) 3x b) 1/3x c) 9x

B) 1/3*x Since the volume triples, and since according to Boyle's Law the volume and pressure are inversely proportional, the pressure will fall by a factor of 3.

Is high a level of oxygen harmful?

High oxygen levels can also have negative physiological effects.

Why does hot air rise?

Hot air rises because the volume of a gas sample at constant pressure increases with increasing temperature. The volume of a gas increases with increasing temperature. Temperature and volume are linearly related.

What is hypoxia?

Hypoxia is oxygen starvation. Mild hypoxia causes dizziness, headache, and shortness of breath. Severe hypoxia which occurs when PO2 drops below 0.1 atm, may cause unconsciousness or even death.

A sample of gas has a mass of 827 mg. Its volume is 0.270 L at a temperature of 88 degrees C and a pressure of 975 mmHg. Find the molar mass. Given: mass = 827 mg; V = .270L; T = 88 C; P = 975 mmHg R = 0.0821 Find: n moles PV = nRT Molar Mass = Mass/moles (n)

Mass = 827 mg; 1 gram = 1000 mg Mass = 827 mg * 1 g/1000 mg = .827 g T = 88 C = 88 + 273.15 = 361.15K P = 975 mmHg; 1 atm = 760 mmHG 975 mmHg * 1 atm/760 mmHg = 1.2829 atm n = PV/RT n = (1.2829 atm * .27 L) / (.0821 * 361.15) n = .011682 mole Molar Mass = .827 g/.011682 = 70.8 g/mol

Given: P1 = 765 mm Hg; V1 = 1.78 L; V2 = 1.25 L Find: P2

P1V1 = P2V2 (765 mm Hg)*(1.78 L) = P2 * 1.25 L 1361.70 mm Hg/L = P2 * 1.25 L P2 = 1089.36 mm Hg = 1.09 X 10^3

Given: n = 0.18 mol; V = 1.2L; T = 298K Find: P R = 0.0821 L * atm/mol * K

PV = nRT P = nRT/V P = [(0.18 mol) * (0.0821) * 298K]/1.2L 3.7 atm

What is the pressure equation?

Pressure = Force/Area

What is partial pressure?

The pressure due to any individual component in a gas mixture. Partial Pressure of Component = Fractional Composition of component X Total Pressure

Given: 0.311 atm Find: mm Hg 1 atm = 760 mm Hg

0.311 atm X 760 mm Hg/1 atm = 236 mm Hg

What are the kinetic molecular theory assumptions?

1. A gas is a collection of particles (molecules or atoms) in constant, straight-line motion. 2. Gas particles do not attract or repel each other - the do not interact. The particles collide with each other and with the surfaces around them, but they bounce back from these collisions like idealized billiard balls. 3. There is a lot of space between gas particles compared with the size of the particles themselves. 4. The average kinetic energy - energy due to motion - of gas particles is proportional to the temperature of the gas in kelvins. This means that as the temperature increases, the particles move faster and therefore have more energy.

What is kinetic molecular theory?

A simple model for gases that predicts the behavior of most gases under many conditions.

What is millimeter of mercury (mm Hg)?

A unit of pressure that originates from the method used to measure pressure with a barometer. Also called a torr. 1 atm = 760 mm Hg

What is nitrogen narcosis?

An increase in nitrogen concentration in bodily tissues and fluids that results in a feeling of drunkenness.

What is extrapolation?

Extending the line on our plot backward from the lowest measured point.

What is molar volume?

Molar volume is the volume occupied by one mole of gas. Under standard temperature and pressure conditions the molar volume of ideal gas is 22.4L Under standard conditions, 1 mol = 22.4 L

How many liters of oxygen gas form when 294 g of KClO3 completely react in this reaction. 2KClO3 (s) ----> 2KCl (s) + 3O2 (g) Assume that the oxygen gas is collected at P = 755 mm Hg and T = 305K Given: 294 g KClO3; P = 755 mm Hg; T = 305K; R = 0.0821 Find: V

PV = nRT Molar Mass of KClO3 = 122.55 1 mole KClO3 = 122.55 grams 2 moles KClO3 = 3 moles O2 1 atm = 760 mm Hg 294 g KClO3 X 1 mole KClO3/122.55 g = 2.399 mol KClO3 2.399 mol KClO3 X 3 mol O2/2 mol KClO3 = 3.599 mol O2 Convert 755 mm Hg to atm P = 755 mm Hg X 1 atm/760 mm Hg = 0.9934 atm V = nRT/P V = (3.599 mol O2) * 0.0821 * 305K/0.9934 atm = 90.7 L

What is oxygen toxicity?

The result of increased oxygen concentration in bodily tissues.

What is Charle's Law?

The volume (V) of a gas and its Kelvin temperature are directly proportional. If two variables are directly proportional, increasing one by some factor increases the other by the same factor. V = Constant * T V/T = Contant V1/T1 = Constant = V2/T1

What is the equation for pressure based on Boyle's Law?

V = 1/P V = Constant/P PV = Constant Constant = P1V1 = P2V2

What is the Combined Gas Law?

A law that combines Boyle's Law and Charle's Law. It is used to calculate how a property of gas (P, V or T) changes when two other properties are changed at the same time. P1 * V1/T1 = P2 * V2/T2 The combined gas law applies only when the amount of gas is constant.

What lowers pressure?

Fewer gas particles lower pressure. Pressure decreases as altitude increases.

What is absolute zero?

The coldest temperature possible. The temperature at which molecular motion stops. Lower temperatures do not exist.

How many liters of oxygen (at STP) are required to form 10.5 g of H2O? 2H2 (g) + O2 (g) ----> 2H2O (g) Given: 10.5 g H2O Find: Liters of O

1 mol O2 = 2 mol H2O 1 mol O2 = 22.4 L Molar Mass H2O = 18.01528 gram 1 mol H2O = 18.01528 g 10.5 g H2O X 1 mol H2O/18.01528 g = 0.5828 mol H2O 0.5828 mol H2O X 1 mol O2/2 mol H2O = 0.2914 mol O2 0.2914 mol O2 X 22.4 L/1 mol O2 = 6.53 L O2

What is atmosphere (atm)?

The average pressure at sea level. 1 atm = 101,325 Pa

A high-performance road bicycle tire is inflated to a total pressure of 125 psi. What is the pressure in millimeters of mercury? 760 mm Hg = 14.7 psi

125 psi X 760 mm Hg/14.7 psi = 6,462 = 6.46 X10^3 mm Hg

Covert a pressure of 173 in Hg into lbs per square inch.

14.7 psi (pounds per sq in) = 29.92 in Hg 173 in Hg x 14.7 psi/29.92 in Hg = 85.0 psi

A snorkeler takes a syringe filled with 16 mL of air from the surface, where the pressure is 1.0 atm, to an unknown depth. The volume of the air in the syringe at this depth is 7.5 mL. What is the pressure at this depth? If the pressure increases by an additional 1 atm for every 10 m of depth, how deep is the snorkeler? Given: V1 = 16 mL; P1 = 1.0 atm; V2 = ?; P2 = 7.5 mL

(16 mL) * (1.0 atm) = (7.5 mL) * V2 16 mL/atm = 7.5 mL * V2 V2 = 2.133 atm 1 atm = 10 meters 2.133 - 1 atm = 1.133 atm 1 atm should be subtracted because you are starting at sea level where pressure is 1 atm and you haven't traveled any distance below water yet. 1.133 atm * 10 = 11 meters

A sample of gas has an initial volume of 3.95 L at a pressure of 705 mm Hg. If the volume of the gas is increased to 5.38 L, what is the pressure? (assume constant temperature) Given: V1 = 3.95L; P1 = 705 mm Hg; V2 = 5.38L Find: P2

(3.95L) * (705 mm Hg) = (5.38L) * P2 2784.75 L/mm Hg = 5.38L * P2 P2 = 518 mm Hg

What are the properties of gases?

1) Gases are compressible. Gases are compressible because the atoms or molecules that compose them have a lot of space between them. 2) Gases assume the shape and volume of their container. Gases assume the shape and volume of their container because gaseous atoms or molecules are in constant, straight-line motion. 3) Gases have low densities in comparison with liquids and solids. Gases have a low density in comparison with solids and liquids because there is so much empty space between the atoms or molecules in a gas.

What are common Units of Pressure?

101,325 Pa (Pascal) = 1 atm (atmosphere) = 760 mm Hg (millimeter of mercury) = 760 torr = 14.7 psi (pounds per square inch) = 29.92 in Hg (inches of mercury in Hg)

How many grams of water form when 1.24 L of H2 gas at STP completely reacts with O2? 2H2 (g) + O2 (g) ----> 2H2O (g) Given: 1.24 L H2 Find: H2O grams

2 mol H2 = 2 mol H2O 1 mol H2 = 22.4 L Molar Mass H2O = 18.01528 gram 1 mol H2O = 18.01528 g 1.24 L H2 X 1 mol H2/22.4 L = 0.05535 mol H2 0.05535 mol H2 X 2 mol H2O/2 mol H2 = 0.05535 mol H2O 0.05535 mol H2O X 18.01528 g H2O/1 mol H2O = 0.997 g H2O

An experiment shows that a 248 mL gas sample has a mass of 0.433 g at a pressure of 745 mmHg and a temperature of 28 degrees C. What is the molar mass of the gas. Given: V = 248mL; mass = 0.433 g; P = 745 mmHg; T = 28 C R = 0.0821 Find: n moles Molar Mass = Mass/Moles (n)

248 mL; 1 L = 1000 mL V = 248 mL * 1 L/1000mL = .248 L P = 745 mmHg; 1 atm = 760 mmHg P = 745 mmHg * 1 atm/760 mmHg = .980263 atm T = 28 C = 28 + 273.15 = 301.15K n = PV/RT n = (.980263 atm * .248L) / (.0821 * 301.15K) n = .00983 moles Molar Mass = .433 g/.00983 = 44.0 g/mol

Convert a pressure of 23.8 in Hg into kilopascals.

29.92 in Hg = 101325 Pa 23.8 in Hg X 101325 Pa/29.92 in Hg = 80599 Pa 80599 Pa / 1000 = 80.6 kilopascals (kPa)

What is the Ideal Gas Law?

A law that combines the four properties of gas - pressure (P), volume (V), temperature (T), and number of moles (n) in a single equation - showing their inter-relatedness. PV = nRT R = ideal gas constant Each of the quantities in the ideal gas law must be expressed in units within R. Pressure - P must be in atmospheres Volume - V must be expressed in liters. Amount of Gas - n must be expressed in moles Temperature - T must be expressed in Kelvins

What is a mixture of gases?

A non-pure gas. It is a mixture if gases. According to the kinetic molecular theory, each of the components in a gas mixture acts independently of the others.

A liquid that is about twice as dense as water is used in a barometer. With this barometer, normal atmospheric pressure would be about: a) 0.38 m b) 1.52 m c) 5.15 m d) 20.6 m

C) 5.15 m Atmospheric pressure will support a column of water 10.3 m in height. if the liquid in a barometer were twice as dense as water, a column of it would be twice as heavy and the pressure is exerted as its base would be twice as great. Therefore, atmospheric pressure would be able to support a column only half as high.

What is the relationship between pressure and volume?

If the temperature and the amount of gas are constant, the pressure of a gas sample increases for a decrease in volume and decreases for an increase in volume. Decrease in Volume = Gas Pressure Increases Increase in Volume = Gas Pressure Decreases

What partial pressure can our lung breathe?

Our lungs have evolved to breathe oxygen at a partial pressure of PO2 = 0.21 atm

Calculate the number of moles of gas in a basketball inflated to a total pressure of 24.2 psi with a volume of 3.2L at 25 degrees C. Given: P = 24.2 psi; V = 3.2L; T = 25 C R = 0.0821 Find = n

P = 24.2 psi ; 14.7 psi = 1 atm 24.2 psi * 1 atm/14.7 psi = 1.646 atm T = 25 C; K = C + 273.15 = 298.15K PV = nRT n = PV/RT n = (1.646 atm) * 3.2L/[0.0821 * 298.15] n = 5.2672/24.478 = .22 mol

A cylinder contains 11.8L of air at a total pressure of 43.2 psi and a temperature of 25 degrees C. How many moles of gas does the cylinder contain? Given: V = 11.8 L; P = 43.2 psi; T = 25 C R = 0.0821 Find : moles n

P = 43.2 psi; 1 atm = 14.7 psi 43.2 psi * 1 atm/14.7 psi = 2.93877 atm T = 25 C = 25 + 273.15 = 298.15K PV = nRT n = PV/RT n = (2.938 atm * 11.8L) / .0821 * 298.15 n = 1.42 moles

How much volume does 0.556 mol of gas occupy when its pressure is 715 mm Hg and its temperature is 58 degrees C? Given: 0.556 mol; P = 715 mm Hg; T = 58 C R = 0.0821 Find: V

P = 715 mmHg; 1 atm = 760 mmHG P = 715 mmHg * 1 atm/760 mmHg = .941 atm T = 58 C; K = 273.15 + C = 331.15 K PV = nRT V = (nRT/P) V = (.556 mol * 0.0821 * 331.15K)/.941 atm V = 16.1 L

A cylinder equipped with a movable piston has an applied pressure of 4.0 atm and a volume of 6.0 L. What is the volume of the cylinder if the applied pressure is decreased to 1.0 atm Given: P1 = 4.0 atm; P2 = 1.0 atm; V1 = 6.0 L

P1V1 = P2V2 (4.0 atm) * (6.0 L) = (1.0 atm) * (V2) 24 atm/L = 1.0 atm * V2 V2 = 24 L

An 8.5 L tire is filled with 0.55 mol of gas at a temperature of 305K. What is the pressure of the gas in the tire? Given: n = 0.55 mol; V = 8.5L; T = 305K R = 0.0821 Find: P

PV = nRT P = (nRT)/V P = (0.55 mol * 0.0821 * 305K) / 8.5L = 1.62 atm

What is the volume occupied by 0.255 mol of helium gas at 1.25 atm and 305K? Given = n = 0.255 mol; P = 1.25 atm; T = 305K R = 0.0821 Find: V

PV = nRT V = (nRT/P) V = (0.255 mol * 0.0821 * 305K) / 1.25 atm V = 5.11 L

Calculate the volume occupied by 0.845 mol of nitrogen gas at a pressure of 1.37 atm and a temperature of 315K. Given: n = 0.845 mol; P = 1.37 atm; T = 315K R = 0.0821 Find: V

PV = nRT V = nRT/P V = [(0.845 mol) * (0.0821) * (315K)]/1.37 V = 16.0 L

Given: m = 0.136 g; V = .112 L; T = 298K; P = 1.06 atm R = 0.0821 Find: Molar mass g/mol PV = nRT Molar Mass = Mass/Moles (n)

PV = nRT n = PV/RT n = (1.06 atm * .112L) / .0821 * 298K n = .00485 moles Molar Mass = .136 g/.00485 = 28.0 g/mol

In this reaction, 4.58L of O2 were formed at 745 mm Hg and 308K. How many grams of Ag2O decomposed? 2Ag2O (s) ----> 4Ag (s) + O2 (g) Given: V = 4.58L; P = 745 mm Hg; T = 308K; R = 0.0821 Find: Grams of Ag2O

PV = nRT 2 moles Ag2O = 1 mol O2 Molar Mass of Ag2O = 231.7358 grams 1 mol Ag2O = 231.7358 grams Convert 745 mm Hg to atm; 1 atm = 760 mm Hg P = 745 mm Hg X 1 atm/760 mm Hg = 0.98026 atm (0.98026) * 4.58L = n * (0.0821) * 308K 4.4896 = n * 25.2868 n = 4.4896/25.2868 = 0.1775 mol O2 0.1775 mol O2 X 2 mol Ag2O/1 mol O2 = 0.355 mol Ag2O 0.355 mol Ag2O X 231.7358 g/1 mol Ag2O = 85.3 g Ag2O

Calculate the partial pressure of oxygen that a diver breathes with a heliox mixture containing 2.0% oxygen at a depth of 100m where the total pressure is 10.0 atm. Given: Mixture contains 2.0% O; Depth = 100 m; Ptot = 10.0 atm Find: Partial pressure of O

Partial Pressure of Component = Fractional Composition of Component X Total Pressure Fractional Comp of O2 = 2/100 = .02 PO2 = .02 * 10.0 = .2 atm

A diver breathing heliox with an oxygen composition of 5.0% wants to adjust the total pressure so that PO2 = 0.21 atm. What must the total pressure be? Given: 5.0% O or 0.05; Partial Pressure = 0.21 atm

Partial Pressure of Component = Fractional Composition of Component X Total Pressure O pressures = 5% = 5/100 = .05 0.21 atm = 0.05 X Total Pressure 0.21 atm/0.05 = 4.2 atm

A gas mixture contains 78% nitrogen and 22% oxygen. If the total pressure is 1.12 atm, what are the partial pressures of each component. Given: 78% N or 0.78; 22% O or .22; Total Pressure = 1.12 atm Find: Partial Pressure of each component

Partial Pressure of N = 0.78 X 1.12 atm = 0.8736 atm Partial Pressure O = 0.22 X 1.12 atm = 0.2464 atm

A heliox deep-sea diving mixture contains 4.0% oxygen and 96.0% helium. What is the partial pressure of oxygen when this mixture is delivered at a total pressure of 8.5 atm. Given: 4.0% O or .04 O; 96% He or .96 He; Total Pressure = 8.5 atm

Partial Pressure of O = .04 X 8.5 atm = 0.34 atm

What is pressure?

Pressure is the result of the constant collisions between the atoms or molecules in a gas and the surfaces around them. The force exerted per unit are by gaseous molecules as they collide with the surfaces around them.

A sample of hydrogen gas is mixed with water vapor. The mixture has a total pressure of 745 torr, and the water vapor has a partial pressure of 24 torr. What is the partial pressure of the hydrogen gas? Given: Ptot = 745 torr; Ph2o = 24 torr Find: PH2

Ptot = PH2O + PH2 PH2 = Ptot - PH2O PH2 = 745 - 24 = 721 torr

A mixture of helium, neon, and argon has a total pressure of 558 mmHg. If the partial pressure of helium is 341 mmHg and the partial pressure of neon is 112 mmHg, what is the partial pressure of argon? Given: Ptot = 558 mmHg; PHe = 341 mmHg; PNe = 112 mmHg Find; PAr

Ptot = PHe + PNe + PAr PAr = Ptot - PHe - PNe PAr = 558 - 341 - 112 = 105 mmHg

A gas mixture contains each gas as the indicated partial pressure. Given: PN2 = 217 torr; PO2 = 106 torr; PHe = 248 torr Find: Ptot

Ptot = PN2 + PO2 + PHe Ptot = 217 + 106 + 248 = 571 torr

What is the standard temperature and pressure (STP)?

Standard temperature and pressure (STP) is when conditions are often assumed in calculations involving gases; T=0 degrees C (273K) and P = 1 atm

A sample of gas has a mass of 0.311 g. Its volume is 0.225L at a temperature of 55 degrees C and a pressure of 886 mmHg. Determine the molar mass. Given: mass = 0.311g; V = .225L; T = 55 C; P = 886 mmHg R = 0.0821 Find: n moles PV = nRT Molar Mass = Mass/Moles (n)

T = 55C = 55 + 273.15 = 328.15K P = 886 mmHg; 1 atm = 760 mmHg 886 mmHg * 1 atm/760 mmHg = 1.165789 atm n = PV/RT n = (1.166 mol * .225L) / (.0821 * 328.15) n = .00974 moles Molar Mass = .311/.00974 = 31.9 g/mol

What is pascal (Pa)?

The SI unit of pressure defined as 1 newton (N) per square meter. 1 Pa = 1 N/m2

What is Boyle's Law?

The volume of a gas and its pressure are inversely proportional. V is inversely proportional to P V = 1/P If two quantities are inversely proportional, increasing one decreases the other.

Given: Total Pressure = 758 mm Hg; Partial Pressure of water = 23.8 mm Hg Find: Partial Pressure of hydrogen gas

Total Pressure = Partial Pressure of hydrogen gas + Partial Pressure of water 758 mm Hg = Partial Pressure of H2 + 23.8 mm Hg Partial Pressure of H2 = 734.2 mm Hg

A volume of gas is confined to a cylinder with a freely movable piston at one end. If you apply enough heat to double the Kelvin temperature of the gas, what happens? a) The volume doubles b) The volume remains the same c) The volume falls to half of the initial volume

V1/T1 = V2/T2 100 mL/200K = V2/400K V2 = 200 mL a) The volume doubles

A gas in a cylinder with a movable piston has an initial volume of 8.2 mL and is heated from 35 degrees C to 155 degrees C. What is the final volume of the gas in mL? Given: V1 = 88.2mL; T1 = 35 degrees C; T2 = 155 degrees C Find: V2

V1/T1 = V2/T2 Gas Law problems must be worked in Kelvins K = C + 2.73.15 T1 = 308.15K; T2 = 428.15K V2 = T2 * V1/T1 = 428.15K * 88.2 mL/308.15K = 123 mL

A balloon with an initial volume of 3.2L at a temperature of 299K is warmed to 376K. What is the volume at 376K? Given: V1 = 3.2L; T1 = 299K; T2 = 376K Find: V2

V1/T1 = V2/T2 V2 = T2 * V1/T1 = 376K * 3.2L/299K = 4.0 L

A sample of gas has a volume of 2.80L at an unknown temperature. When the sample is submerged in ice water at t = 0 degrees C, its volume decreases to 2.57L. What was the initial temperature (in Kelvins and in Celsius)? Assume a constant pressure. Given: V1 = 2.80L; T2 = 0 degree C; V2 = 2.57L Find: T1

V1/T1 = V2/T2 T2 = 273.15 K Gas Law problems must be worked in Kelvins K = C + 2.73.15 T1 = V1 * T2/V2 T1= (2.80L) * 273.15K/2.57L = 297K 24 degrees C

A chemical reaction occurring in a cylinder equipped with a movable piston produces 0.58 mol of a gaseous product. if the cylinder contained 0.11 mol of gas before the reaction and had an initial volume of 2.1 L, what was its volume after the reaction? Given: n1 = 0.11 mol; n2 = (0.58 mol + .11 mol) = .69 mol; V1 = 2.1L Find: V2

V1/n1 = V2/n2 V2 = n2 * (V1/n1) V2 = 0.69 mol * (2.1 L/ 0.11 mol) = 13 L

A 4.8L sample of Helium gas contains 0.22 mol of helium. How many additional moles of helium gas should you add to the sample to obtain a volume of 6.4L? Assume constant temperature and pressure. Given: V1 = 4.8L; 0.22 moles of He; V2 = 6.4L Find: moles of He

V1/n1 = V2/n2 n2 = V2 * (n1/V1) n2 = 6.4L * (0.22 moles He/4.8 L) = 0.29 mol of He 0.29 - .22 = 0.07 mole of He should be added

A 0.12 mol sample of nitrogen gas occupies a volume of 2.55L. What is the volume of 0.32 mol of nitrogen gas under the same conditions? Given: n1 = 0.12 mol; V1 = 2.55L; n2 = 0.32 mol Find = V2

V1/n2 = V2/n2 V2 = n2 * (V1/n1) V2 = 0.32 mol * (2.55L/0.12 mol) = 6.8 L

Given = P1 = 755 mm Hg; V1 = 3.65L; T1 = 302K; T2 = 291K; P2 = 687 mm Hg Find V2

V2 = T2/P2 * [P1*V1/T1] V2 = 291K/687 mm Hg * [755 mm Hg * 3.65L/302K] V2 = .423580 K/mm Hg *[9.125 mmHg/LK V2 = 3.87 L

What is vapor pressure?

Vapor pressure is the partial pressure of a vapor in dynamic equilibrium with its liquid.

A sample of gas has an initial volume of 158 mL at a pressure of 735 mm Hg and a temperature of 34 degrees C. If a gas is compressed to a volume of 108 mL and heated to a temperature of 85 degrees C. What is the final pressure in mm Hg? Given: V1 = 158 mL; P1 = 735 mm Hg; T1 34 C or 307.15K V2 = 108 mL; T2 = 85 C or 358.15K Find: P2

[P1 * V1]/T1 = [P2 * V2]/T2 P2 = T2/V2 * [P1 * V1/T1] P2 = (358.15K/108 mL) * [735 mm Hg * 158 mL/307.15K] P2 = 3.3162 * [378.088] = 1254 mm Hg = 1.25 X 10^3 mm Hg

A volume of gas is confined to a container. If you apply enough heat to double the Kelvin temperature of the gas and expand the size of the container to double it initial volume, what happens to the pressure? a) The pressure doubles b) The pressure falls to half of its initial volume c) The pressure is the same as its initial value

[P1 * V1]/T1 = [P2 * V2]/T2 T1 = 100K; T2 = 200K V = 20; V2 = 40 P1 = 10 P2 = (T2/V2) * [P1 * V1]/T1 P2 = (200/40) * [10 * 20]/100 P2 = 5 * 2 = 10 C) The pressure is the same as its initial value

A balloon has a volume of 3.7L at a pressure of 1.1 atm and a temperature of 30 degrees C. If the balloon is submerged in water to a depth where the pressure is 4.7 atm and the temperature is 15 degrees C. What will its volume be? Given: V1 = 3.7L; P1 = 1.1 atm; T1 = 30 C or 303.15K P2 = 4.7 atm; T2 = 15 C or 288.15K Find: V2

[P1 * V1]/T1 = [P2 * V2]/T2 V2 = T2/P2 * [P1 * V1]/T1 V2 = 288.15K/4.7 atm * [1.1 atm * 3.7L]/303.15K V2 = 61.308 * .01342 = .823 L

Find the pressure in mmHg of a 0.133 g sample of helium gas at 32 degrees C contained in a 648 mL container. Given: n = 0.133 g of He; T = 32 C; V = 648 mL R = 0.0821 Find P

n = 0.133 g He 1 mol He = 4.0026 n = .133 g * 1 mol He/4.0026 g = .03325 mol of He T = 32 C = 32 + 273.15 = 305.15K V = 648 mL 1 L = 1000 mL 648 mL * 1 L/1000 mL = .648 L PV = nRT P = (nRT)/V P = (.03325 mol * .0821 * 305.15 K)/.648 L P = 1.2855 atm 1 atm = 760 mmHg 1.28 atm * 760 mmHg/1 atm = 977 mm Hg

What is the equation for n?

n = PV/RT P = Pressure V = Volume T = Temperature R = Ideal Gas Constant = 0.0821

How many moles of NH3 are formed by the complete reaction of 2.5L of hydrogen at 381K an 1.32 atm. Assume there is more than enough N2 Balanced Equation: 3H2 (g) + N2 (g) ---> 2NH3 (g) Given: V = 2.5L; T = 381K; P = 1.32 atm (H2) R = 0.0821 Find: mol NH3

n = PV/RT or PV = nRT 3 moles H2 = 2 moles NH3 n = 1.32 atm * 2.5L / 0.0821 * 381K = 3.3 / 31.2801 = 0.1055 mol H2 n = 0.1055 mol H2 Convert mol H2 to mol NH3 0.1055 mol H2 X 2 mol NH3/3 mol H2 = 0.070 mol NH3


Conjuntos de estudio relacionados

CSUF GEOG332 US Geography Smith - Exam 2

View Set

Ch 10 Laws Governing Access to Foreign Markets

View Set

Intermediate accounting 2 Chapter 15 & 16

View Set

unit 1 2 Introduction to Basic Essay Formatting

View Set

POI: (43) Tumörimmunologi - Essand

View Set