Gas Law Practice Exam

Which of the following conditions would be most likely to cause the ideal gas laws to fail? I. High pressure II. High temperature III. Largevolume

A.I only-- Gas laws don't work out as well when the molecules are pushed too close together, causing them "notice" each other and "feel the stickiness."

The temperature of a sample of an ideal gas confined in a 2.0 L container was raised from 27oC to 77oC. If the initial pressure of the gas was 1,200 mmHg, what was the final pressure of the gas?

B. 1,400 mmHg--Using the combined gas law with moles and volume constant. P1/T1 = P2/T1 Substitute (350)(1200) = P2 and look for common factors.

When 4.0 moles of oxygen are confined in a 24-liter vessel at 176oC, the pressure is 6.0 atm. If the oxygen is allowed to expand isothermally until it occupies 36 liters, what will be the new pressure?

C.4 atm-- The term, isothermally, means occurs at the same temperature. Thus this is a combined gas law problem with equal moles and equal temperature. PV = P V Solve for P2 and substitute and look for common factors. P1V1/V2 = P2 (6)(24)/36 = P2

A mixture of gases contains 1.5 moles of oxygen, 3.0 moles of nitrogen, and 0.5 mole of water vapor. If the total pressure is 700 mmHg, what is the partial pressure of the nitrogen gas?

E. 420 mmHg--The is the same problem as the previous problem, solve for the mole fraction of nitrogen, (3)/(5) 700 . You will likely find it easiest to divide the 5 into the 700 to get 140, then triple that to get 420

Liquid nitrogen has a boiling point of -196oC. This corresponds to

b.77K--K = oC + 273

A hydrocarbon gas with the empirical formula CH2 has a density of 1.3 g/L at 0oC and 1.00 atm. A possible formula for the hydrocarbon is:

b.C2H4- In this problem, we need a molar mass to determine the molecular formula. Notice that the conditions in the problem, 0oC and 1 atm are STP conditions, and this allows us to use 22.4 L/mol. 1.3g/1L × 22.4/1mol=29 molar mass, Of the options given, the only formula with a molar mass of ~29 is (b), C2H4 with a molar mass of 28.

1.5 atm is the same pressure as

A.1140 mmHg-- atm is changed to mmHg by multiplying by 760.

A 1.1 L container will hold about 1.6 g of which of the following gases at 0oC and 1 atm?

B. O2 Don't miss the fact that this problem is at STP. To figure out which gas, you need to calculate a molar mass. 1.6g/1.1L × 22.4L/1mol . This may look "messy" but the 1.1 divide into the 22.4 ~20 times, and thus 1.6 × 20 is just over 30, thus oxygen.

A sealed container containing 8.0 grams of oxygen gas and 7.0 g of nitrogen gas is kept at a constant temperature and pressure. Which of the following is true?

B. The volume occupied by oxygen is equal to the volume occupied by nitrogen.-- When give masses it is important to convert to moles so you can compare the quantities of gas molecules. 8 g of O2 is 0.25 mol and 7 g of N2 is also 0.25 mol, Thus at constant temperature and pressure, for equal moles of gas, the volume would remain constant.

What is the total volume of products formed at STP when 1.2 g of carbon is burned?

B.2L-- Write a balanced equation: C + O2 → CO2 1.2g × 1mol/12g × 1C/1CO2 × 22.4L/1mol = 2L

A mixture of helium and neon gases has a total pressure of 1.2 atm. if the mixture contains twice as many moles of helium as neon, what is the partial pressure due to neon?

C. 0.4 atm--The information tells us that the neon is 1⁄3 of the total amount of gas, thus (1)/3 1.2 = 0.4

The reaction below takes place in a closed, rigid vessel. The initial pressure of N2(g) is 1.0 atm, and that of O2(g) is 1.5 atm. No N2O4(g) is initially present. The experiment is carried out at a constant temperature. What is the total pressure in the container when the partial pressure of N2O4 reaches 0.75 atm? N2 + 2O2 → N2O4

C. 1.0 atm--Set up a ICE box. When the temp and volume are constant, the pressure of a gas is proportional to its number of moles, so you can treat the pressure values, the same way you would treat mole values. Set up a rice box as shown, which allows you to calculate the pressure at the end using Dalton's Law of Partial Pressures; Ptotal = P1 + P2 + P3 + ... In the ICE box shown below, you can see the pressure of the nitrogen and oxygen given in the I row, and the problem tells us there is no product. The problem also tells us there is 0.75 of the product at the end. This lets us calculate how much oxygen was used up to produce the 0.75 atm of product: 0.75atm × 2O2/1N204 = 1.5atm for O2 and how much nitrogen was used to produce the 0.75 atm of product: 0.75atm × 1N2/1N2O4=0.75atm for N2 Put these in the change row and then do the subtraction to find the End row 0.75+0.25=1 ATM

When a sample of carbon dioxide gas in a closed container of constant volume at 0.5 atm and 200 K is heated until its temperature reaches 400 K, its new pressure is closest to

C. 1.5 atm--The should appear to be a "before and after" problem, thus use the combined gas law without n, since the container is sealed, and without volume since you are told that there is a constant volume. P1/T1 = P2/T2 Substitute 0.5/200 = P2/400 and solve.

A sample of 18.0 g of aluminum metal is added to excess hydrochloric acid. The volume of hydrogen gas produced at 0.0oC and 1 atm pressure is approximately

C. 22 L--Again, you must write a balanced equation: 2Al + 6HCl → 2AlCl3 + 3H2 Convert the 18 g of Al to moles 2mol×3H2/2AL ×22.4L/1mol=~22L

A flask with a pressure of 2000 mmHg contains 6 mol of helium with a partial pressure of 1500 mmHg. The remaining gas is hydrogen, what is the mass of hydrogen in the flask?

C. 4.0g--We know that each gas causes a portion of the total pressure. Thus we can make a proportion using moles and pressures. N1/P1=Ntotal/Ptotal to solve for the total number of moles Ptotal n1/P1 = ntotal (2000)(6)/(1500) = Ntotal Thus the total number of moles is 8, and 1⁄4 of that is 2 moles of hydrogen with a molar mass of 2 g/mol

An ideal gas fills a ballon at a temperature of 27oC and 1 atm pressure. By what factor will the volume of the balloon change if the gas in the balloon is heated to 127oC at constant pressure?

C. 4/3--Use the combined gas law V1/T1 = V2/T2 and be sure and convert to Kelvin. Solve T2V1/T1 = V2 and substitute 400V1/300 = V

A given mass of a gas occupies 5.00 L at 65 °C and 480 mmHg. What is the volume of the gas at 630 mmHg and 85 °C

C. 5.00*358/338*480/630-- Rearrange the combined gas law and change to Kelvin, again assuming the container is sealed and moles remains constant. P1V1/T1 = P2V2/T2 , thus V1 T2/T1 P1/P2=V2

The pressure of 4.0 L of an ideal gas in a flexible container is decreased to one-third of its original pressure and its absolute temperature is decreased by one-half. The volume then is

C. 6.0L-- Use the combined gas law with n removed because without any other information, you can assume the container is sealed. P1 V1/T1= P2 V2/T2 though you may find it far less Substitute with pressure and temp with values of 1 (1)(4L)/1= 1/3V2/1/2 complicated by picking the original pressure as 3 and temperature as 2 (3)(4L)/2 = (1)V2/1 and then solve.

A 14.0 gram sample of an unknown gas occupies 11.2 liters at standard temperature and pressure. Which of the following could be the identity of the gas?

C. I and III only--Take the easy route by realizing this problem is at STP, and you know that the volume of a mole of any gas at STP is 22.4 L/ mol. Thus 11.2 L is a 1⁄2 mol of gas. (14g)/0.5 mol = 28g / mol

A given mass of gas in a rigid container is heated from 100 °C to 300 °C. Which of the following best describes what will happen to the pressure of the gas? The pressure will

C.increase by a factor less than three-- Because of the combined gas law, P1/T1 = P2/T2 , which you can rearrange, T2 P1/T1 = P2 , thus you may be tempted to say the pressure would be triple, but you must change to Kelvin temperature (573)P1/373 which does not triple the pressure.

The coldest possible temperature of any substance is

D. -273 K--When you cool a substance down, the molecules slow down, and molecular motion would stop at absolute zero, 0 K or −273oC. When this occurs, there is no way to slow the molecules further, thus there is no way to cool the temperature further, making absolute zero the coldest possible temperature.

A real gas would act most ideal at

D. 0.5 atm and 546 K--Gases follow the gas laws best (most mathematically correct) when they are not be pushed to turn in to a liquid. Gases will liquify at high pressure and low temperature, thus gases behave most ideally at lower pressure and higher temperatures.

A gas sample contains 0.1 mole of oxygen and 0.4 mole of nitrogen. If the sample is at standard temperature and pressure, what is the partial pressure due to nitrogen?

D. 0.8 atm--The mole fraction (symbolized N1) of a gas causes that same fraction of the total pressure. N1Ptotal = P1 The mole fraction of the nitrogen is (0.4)/0.5 Thus (4)/5 1 = 8

A sealed flask at 20oC contains 1 molecule of carbon dioxide, CO2 for every 3 atoms of helium, He. If the total pressure is 800 mmHg, the partial pressure of helium is

D. 600 MMHG--The helium must be 3⁄4 of the total pressure.

Consider the reaction below, in which 6 atm of ammonium nitrite is added to an evacuated flask with a catalyst and then heated. NH4NO2(g) → N2(g) + 2 H2O(g) At equilibrium the total pressure is 14 atm. Calculate the partial pressure of the water vapor at equilibrium

D. 8.0 atm--It's a good idea to use a RICE box to solve this problem. The volume is constant, and we can assume that at equilibrium (the end) that the temperature will return to the starting temperature, thus T is constant, which allows us to substitute with pressure values. We know that the pressure at the end must add up to 14. Thus (6 − x) + x + x = 14 and solve for x = 4 but the question asks for the pressure of water vapor, which is 2x.

Consider the combustion of 6.0 g of ethane. What volume of carbon dioxide will be formed at STP?

D. 9.0 L--This is a stoichiometry problem that requires a balanced equation. Ethane is C2H6 thus C2H6 + 7/2O2 → 2CO2 + 3H2O Convert the ethane to moles, but keep it as a fraction, 1⁄5 Thus 1/5 × 2CO2/1C2H6 × 22.4 = While this math may not look so smooth, consider the 22.4 to be ~20, to get an answer of ~8 and survey the answers to see that a value of 9 L compared to the rest of the choices, is close enough to select and move on.

Equal numbers of moles of CO2(g), N2(g), and NH3(g) are placed in a sealed vessel at room temperature. If the vessel has a pinhole-size leak, which of the following will be true after some of the gas mixture has effused?

D. The mole fraction of CO2 in the sample will increase.--As the gases effuse out of the vessel, the gases that have a slowest velocity will leak out more slowly, thus carbon dioxide with the largest molar mass will go out the slowest, as the other gases move out faster, and the carbon dioxide's moles proportional to the other gases will become higher than at the start. The other statements are flat out false; N2 is not the lightest gas, NH3 is the lightest. The gases will effuse at different rates because they all have different molar masses.

Real gases vary from the ideal gas laws gases at conditions of

D. low temp and high pressure--Remember that the gas laws "work best" and a gas behaves most ideally when their molecules are far apart and moving very fast. Gases vary from the ideal when their molecules begin to slow down enough, lower temps, and get pushed close together enough, high pressure, that they start to feel their intermolecular attractions for one another. This makes them "sticky" to each other and eventually they will turn to a liquid. Even before condensation can occur, the gas law proportions will not mathematically work out so well.

Nitrogen gas was collected over water at 25oC. If the vapor pressure of water at 25oC is 23 mmHg, and the total pressure in the container is measured at 781 mmHg, what is the partial pressure of the nitrogen gas?

E. 758mmHg--Just like our lab work, the pressure for any gas collected over water is caused by the gas and the water vapor pressure. A simple subtraction allows the calculation of the gas itself. 781 − 23 = 758.

Chlorine gas and fluorine gas will combine to form one gaseous product. One L of Cl2 reacts with 3 L of F2 to produce 2 L of product. assuming constant temperature and pressure condition, what is the formula of the product?

E. ClF3-- Since the reaction is at constant temp and pressure, you can write an equation using the volumes as moles. Knowing the stoichiometry of the reactants and products will tell you what the formula must be. Cl2 + 3F2 → 2 Cl?F? thus ClF3

A gaseous mixture at a constant temperature contains O2, CO2, and He. Which of the following lists the three gases in order of increasing average molecular speeds?

E.CO2,O2,CO2-- As you must know by now, molecular speed of gases is inversely proportional to their molar mass, thus list the gases from largest MM to smallest MM.

Which of the following assumption(s) is (are) valid based on kinetic molecular theory. I. Gas molecules have negligible volume. II. Gas molecules have no attractive forces on each other. III. The temperature of a gas is directly proportional to its kinetic energy.

E.I, II,and III-- These are the three most important assumptions of the KMT.

Two flexible containers for gases are at the same temperature and pressure. One holds 14 g of nitrogen and the other holds 22 g of carbon dioxide. Which of the following statements about these gas samples is true?

a. The volume of the carbon dioxide container is the same as the volume of the nitrogen container.---It is important to convert to moles to consider the situation presented. 14 g of N2 is 0.5 mol and 22 g of CO2 is 0.5 mol. You are meant to realize that two containers with the same number of moles at the same temperature and same pressure will have to be the same volume.

A 0.33 mole sample of CaCO3(s) is placed in a 1 L evacuated flask, which is then sealed and heated. The CaCO3(s) decomposes completely according to the balanced equation below. The total pressure in the flask, measured at 300 K is closest to which of the following? (The gas constant, R = 0.082 L atm/mol K) CaCO3(s) → CaO(s) + CO2(g)

c. 8.1--The word "evacuated in the problem means emptied out. Thus there is no carbon dioxide at the start. The simple stoichiometry tells us 0.33 mol of calcium carbonate will form 0.33 mol of carbon dioxide. Then use the ideal gas law to calculate pressure (0.33mol)⎛ 0.082atmiL⎞ (300K)/1L=8.1 calculate pressure. PV = nRT

A sample of 0.010 mole of nitrogen dioxide gas is confined at 127oC and 2.5 atmospheres. What would be the pressure of this sample at 27oC and the same volume?

e. 1.88-This problem is a "before and after" problem, which means you need to break out the combined gas law. Gay-Lussacs law p1/t1=p2/t2. Reread the problem and realize that n and V remain constant, thus the problem reduces to 2.5/400 = P2/300 . you should realize you are looking for 3⁄4 of 2.5, half of 2.5 is 1.25, and it must be larger than this, thus the only option is (e).

A gas sample with a mass of 10 grams occupies 6.0 liters and exerts a pressure of 2.0 atm at a temperature of 26oC. Which of the following expressions is equal to the molecular mass of the gas?

A. (10)(0.0821)(299)/(2.0)(6.0) g*mol-1-- You can revise the ideal gas law PV = nRT substituting n = m/mm , OR use the Molar Mass Kitty Kat! MM = DRT/P , but remember that D = m/V , so you can substitute it back in MM = mRT/VP . Then substitute in the data.

A 7 Liter container will hold about 12 g of which of the following gases at 0oC and 1 atm?

A. F2-- Don't miss the fact that this problem is also at STP, thus 7 L is ~1⁄3 of a liter. Calculate molar mass MM= 12g/1/3 mol =36g/mol

A gaseous mixture of oxygen and nitrogen is maintained at a constant temperature. Which of the following MUST be true regarding the two gases?

A. Their average kinetic energies will be the same.-- If substances are at the same temperature, they must have the same amount of kinetic energy.

If 2.0 moles of gas in a sealed glass flask is heated from 25oC to 50oC. Select the conditions that are true.

A. increases, increases, stays the same-- When temperature increases, kinetic energy increases, which in turn causes pressure to increase since the flask has constant volume. The number of moles is constant because the flask is sealed.

A gas sample is confined in a 5-liter container. Which of the following will occur if the temperature of the container is increased?

C. I and II only When you increase the temperature of a sealed rigid container, the mass nor the volume will change, thus the density will not change. Kinetic energy will increase, which in turn increased the pressure.

An ideal gas contained in 5.0 liter chamber at a temperature of 37oC. If the gas exerts a pressure of 2.0 atm on the walls of the chamber, which of the following expressions is equal to the number of moles of the gas?

E. (2.0)(5.0)/(0.082)(310)mol-- Use the ideal gas law and solve for n. n = PV/RT Substitute in the values given in the problem.

A 5.0 L sample of gas is collected at 400. mmHg at 727oC. What is the volume if the temperature were cooled to 77oC and the pressure increased to 700. mmHg?

E. 1.0L-- Use the combined gas law P1V1 = P2V2 , remember to convert to Kelvin, and assume moles are constant. (400)(5)/(1000) = (V2)(700)/(350) Rearrange and look for common factors (350)(400)(5)/(700)(1000) = V2

For an ideal gas, which pair of variables are inversely proportional to each other (if all other factors remain constant)?

a. P, V--This is Boyle's Law, part of the combined gas law.