chem4a chapter 10

A container initially holds 5.67×10−2mol of propane and has a volume of V1. The volume of the container was increased by adding an additonal 2.95×10−2mol of propane to the container, so that the container has a final volume of 1.93L. If the temperature and pressure are constant, what was the initial volume of the container, V1?

1.27L First, calculate the final number of moles of propane (n2) in the container. n2=n1+nadded=5.67×10^−2mol+2.95×10^−2mol=8.62×10^−2mol Rearrange Avogadro's law to solve for V1. V1=V2×n1/n2 Substitute the known values of n1, n2, and V2, V1=1.93L×5.67×10^−2mol/8.62×10^−2mol=1.27L

A container initially holds 1.24mol of hydrogen gas and has a volume of 27.8L. Hydrogen gas was added to the container, and the final volume increased to 40.6L while the temperature and pressure remained constant. How many moles of hydrogen gas are in the container?

1.81mol Rearrange Avogadro's law to solve for n2. n2=V2×n1/V1 Substitute the known values for n1, V1, and V2. n2=40.6L×1.24mol/27.8L=1.81mol

A container with 0.255mol of helium gas has a volume of 3.42L. What will the volume be if an additional 0.510mol of gas is added into the container, assuming the temperature and pressure remain constant?

10.3L.

A container with 0.255mol of helium gas has a volume of 3.42L. What will the volume be if an additional 0.510mol of gas is added into the container, assuming the temperature and pressure remain constant?

10.3L. V1/n1=V2/n2 V2=V1×n2/n1 V2=3.42L×0.765/mol0.255mol V2=10.26L The final volume rounded to three significant figures is 10.3L.

What will be the mass of one mole of carbon atoms?

12 g

Helium is used to inflate a weather balloon from an initial volume of 2,280L to a final volume of 5,710L under constant temperature and pressure. The final volume contains 255mol of helium. How many moles of helium were added to the balloon?

153mol of helium gas must have been added to the balloon. 2,280L×255mol5,710L=101.82mol Lastly, to find the amount of gas that must have been added, calculate the difference between n2 and n1. nadded=n2−n1=255mol−101.82mol=153.18mol Therefore, after rounding this value to three significant figures, we find that approximately 153mol of helium gas must have been added to the balloon.

A sample of a gas has a volume of 30.0 mL at a pressure of 6.50 psi. Determine the volume of the gas in milliliters at a pressure of 11.0 psi.

17.727 mL V2=P1V1/P2 Next, substitute in the known values and solve. V2=(6.50 psi)(30.0 mL)/11.0 psi =17.727 mL

Which of the following represents standard pressure? 1atm 101.325atm 1kPa 101.325kPa

1atm 101.325kPa Standard pressure is defined as 1atm or 101.325kPa.

Radon is a radioactive noble gas that can sometimes be found in unventilated basements. A sample of 1.35×10−4mol of radon gas is held in a container with a volume of 3.03mL. A quantity of radon gas is added to the container, which is then found to have a volume of 7.79mL at the same temperature and pressure. How many moles of radon were added to the container?

2.12×10^−4mol First, find the final number of moles of radon in the container after the addition by rearranging Avogadro's law to solve for n2. n2=V2×n1V1 Substitute the known values ofn1, V1, and V2. n2=7.79mL×1.35×10−4mol3.03mL=3.47×10−4mol Find the difference between the final number of moles (n2) and the initial number of moles (n1). n2−n1=3.47×10−4mol−1.35×10−4mol=2.12×10^−4mol

A container of acetylene gas (C2H2) holds 3.86mol of acetylene and has a volume of 39.2L. If there were a leak and the volume were to decrease to 17.3L at the same pressure and temperature, how many moles of acetylene would have been lost from the container?

2.16mol n2=V2×n1V1 Substitute in the known values for V1, V2, and n1. n2=17.3L×3.86mol/39.2L=1.70mol Find the difference between the initial number of moles (n1) and the final number of moles (n2) n1−n2=3.86mol−1.70mol=2.16mol

A container with a volume of 623L and holds nitrogen gas with a pressure of 111atm. How many moles of gas does it contain at 12.00∘C? Assume ideal gas behavior.

2.96×103 mol. Starting from the ideal gas law PV=nRT isolate n and substitute in the given conditions. n=PVRT=(111 atm)(623 L)/(0.08206Latmmol−1K−1)(273.15+12.00) K≈2955.3 mol=2.9553×103mol

A sample of gas has a volume of 15.0 mL at a pressure of 13.0 psi. Determine the pressure of the gas at a volume of 7.5 mL.

26 psi P1V1=P2V2 13.0 psi×15.0 mL=P2×7.5 mL Solving: P2=13.0 psi×15.0 mL/7.5 mL=26 psi

A container of argon gas has a volume of 67.3mL. After more argon is added, the container holds 4.40×10−3mol of argon and the volume of the container expands to 98.5mL under conditions of constant temperature and pressure. How many moles of argon were initially in the container?

3.01×10−3mol

A container holds 7.32g of nitrous oxide. After the addition of 14.1g of nitrous oxide, the container expands to a volume of 10.9L under constant temperature and pressure. What was the initial volume of the container? The molecular formula of nitrous oxide is N2O and its molar mass is 44.01g/mol. Give your answer in three significant figures.

3.72L Substitute in the known values for n1, n2, and V2. V1=10.9L×0.166mol0.486mol=3.72L

What is the value for a temperature of 373.15K in ∘C?

373.15K equals 100∘C.

A researcher studying ozone (O3) fills a plastic bag with ozone and finds that it contains 2.01g of ozone and has a volume of 0.938L. The bag is then filled with additional ozone to make a volume of 2.34L at the same temperature and pressure. How many grams of ozone are now in the bag? The molar mass of ozone is 47.997 g/mol.

5.01g of ozone First, calculate the number of moles of oxygen in the bag initially. n1=2.01g/47.997g/mol=0.0419mol Next, rearrange Avogadro's law to solve for n2. n2=V2×n1/V1 Substitute in the known values of n1, V1, and V2. n2=2.34L×0.0419mol/0.938L=0.104mol Finally, convert the moles to grams using the molar mass. mozone=0.104mol×47.997g/mol=5.01g So there are 5.01g of ozone in the bag.

Which of the following is equivalent to a mole of oxygen atoms?

6.022×10^23 oxygen atoms

A container that holds 0.152mol of xenon gas has a volume of 3.41L. What will the volume be if an additional 0.190mol of xenon is added to the container and the temperature and pressure are kept constant?

7.67L.

A vapor has a mass of 0.846 g, a volume of 354 mL, a pressure of 752 torr, and a temperature of 100.00∘C. What is the molar mass of the vapor? Use R=0.08206L atm/mol K for the gas constant. Use −273.15∘C for absolute zero.

74.0gmol Mm=mRT/PV 752 torr×1atm/760 torr=0.98947 atm TK=100.00∘C+273.15=373.15 K Mm=(0.846 g)(0.08206L*atm/molK)(373.15 K)/(0.98947 atm)×(0.354 L)= 73.957g/mol

A container holds 3.41×10−3mol of carbon dioxide (CO2). After the addition of 8.41×10−4mol of carbon dioxide, the volume of the container increases to 95.2mL, with the temperature and pressure remaining constant. What was the initial volume of the container? Give your answer in three significant figures.

76.4mL. First, calculate the final number of moles of carbon dioxide (n2) in the container. n2=n1+nadded=3.41×10−3mol+8.41×10−4mol=4.25×10−3mol Rearrange Avogadro's law to solve for V1. V1=V2×n1n2

The ideal gas law displays the relationship in which of the following?

Amontons's law Charles's law Boyle's law Avogadro's law The ideal gas law (PV=nRT) incorporates all of the following relationships: Boyle's Law (P×V), Amontons's Law (PT), Charles's Law (VT), and Avogadro's law (Vn). These laws are combined to determine the relationship between pressure, volume, temperature, and number of moles in the ideal gas law.

Molar mass (M) is equal to the following, where m is the mass in grams and n is the amount in moles:

By definition, the molar mass of a substance is equal to the mass in grams exhibited per mole of that substance, or m/n.

Standard Conditions of Temperature and Pressure

Chemists sometimes make comparisons against a standard temperature and pressure (STP) for reporting properties of gases: 273.15 K and 1 atm (101.325 kPa). At STP, one mole of an ideal gas has a volume of about 22.4 L; this is referred to as the standard molar volume

What are the units for the ideal gas constant R when the temperature is in kelvin, the volume is in liters, and the pressure is in atmospheres?

L*atm/mol−1K−1 The units of R must be consistent with the units of PVnT, so you can see what they must be by substituting the units for each quantity. units of R=units of PVnT=atm L*mol K=L*atm/mol−1K−1

A container holds 17 mol of gas at a temperature of 67.00∘C. If the volume is 88.9 L, what is the pressure of the gas? Use −273.15∘C for absolute zero. Use R=0.08206L atm/mol K for the ideal gas constant.

P=nRT/V=(17mol)(0.08206L atm/mol K)(340.15K)/(88.9L)=5.34atm Therefore, after rounding the answer to two significant figures, we find that the pressure is 5.3atm.

Write the unit abbreviation, do NOT write the full name of the unit.

Pa The SI unit of pressure is the pascal, abbreviated as Pa. Standard pressure is defined as 1atm or 1.01325×105Pa in SI units.

A container with a volume of 25.5L holds 1.050mol of oxygen gas (O2), whose molar mass is 31.998gmol. What is the volume if 7.21g of oxygen gas is removed from the container, assuming the pressure and temperature remain constant?

So the volume is 20.0L.

Give the numeric value for the standard pressure in kPa with an accuracy of one digit after the decimal point.

Standard pressure is defined as 1atm or 101.325kPa.

Which of the following represents standard temperature? 273.15K 298.15K 25∘C 0∘C

Standard temperature is defined as 273.15K or 0∘C.

What is true for ideal gases at STP conditions? Select all that apply. The standard molar volume is 22.4L. All gases have the same mass. All gases have the same reactivity. One mole of any ideal gas has the same volume.

The standard molar volume is 22.4L. One mole of any ideal gas has the same volume. Chemists sometimes make comparisons against a standard temperature and pressure (STP) for reporting properties of gases. At STP, one mole of an ideal gas has a volume of 22.4L. This is referred to as the standard molar volume.

Which is a correct way of stating Boyle's law? P=k×1V PV=k P1V1=P2V2 all of the above

all the above

Which is a correct way of stating Boyle's law? P=k×1V, where k is a constant. PV=k, where k is a constant. P1V1=P2V2, where the indices 1 and 2 corresponds to different states of the same gas sample. all of the above

all the above

Which of the following represents STP? Select all that apply. 1atm and 273.15K 101.325kPa and 273.15K 1atm and 0∘C 101.325kPa and 0∘C Answer Explan

all the above

Assuming the temperature and the amount of gas are held constant, a plot of volume vs. pressure will give a:

hyperbola If we plot pressure versus volume, we obtain a hyperbola. This is because pressure and volume exhibit inverse proportionality; increasing the pressure results in a decrease of the volume of the gas. Mathematically, this can be written as follows:

Give the numeric value in units of Lmol for the standard molar volume of an ideal gas at STP with the accuracy of one digit after the decimal point.

ideal gas has a volume of 22.4L/mol.

Calculating the molar mass of a gas can be useful because it will allow us to:

identify the gas Using the molar mass, we can usually determine the precise identity of the gas, as most common gases have a molar mass that is not shared with other gases.

What does STP mean?

standard temperature and pressure

Avogadro's Law states that for an ideal gas:

volume and moles are directly proportional If V1/n1=V2/n2, then volume and moles must be directly proportional.

Under what condition does the volume of the ideal gas approach zero?

when the temperature approaches 0 K

A sample of 5.6 L of a gas is at a pressure of 1.5 atm. If the volume of the gas is compressed to 4.8 L, what will the new pressure be?

x=1.8 atm Boyle's law states that P1V1=P2V2. Substitute in the numbers that are given. (1.5 atm)(5.6 L)=(x)(4.8 L) Finally, solve for the missing pressure, rounding to two significant figures. x=1.8 atm

Which of the following correctly represents the standard conditions for temperature and pressure (STP)? zero degrees Celsius and one atmosphere room temperature and one atmosphere zero degrees Celsius and 1kPa room temperature and 100kPa

zero degrees Celsius and one atmosphere

Calculate the pressure in bar of 2,520 moles of hydrogen gas (H2(g)) stored at 27∘C in the 180 L storage tank of modern hydrogen-powered car. Use 8.314kPa L/mol K for the ideal gas constant. Recall that 100kPa=1bar.

· PV=nRT · P=nRT/V · =(2520mol)(8.314kPaLmol-1K-1)(300K)/180L≈34,919 kPa · Finally, convert the pressure to bars and round to two significant figures. · 34,919 kPa×1 bar100 kPa≈350 bar

When filled with air, a typical scuba tank with a volume of 13.2 L has a pressure of 153 atm (Figure 1). If the water temperature is 27∘C, how many liters of air will such a tank provide to a diver's lungs at a depth of approximately 70 feet in the ocean where the pressure is 3.13 atm?

·667 L P1V1/T1=P2V2/T2 · Solving for V2 · V2=P1V1T2/T1P2 · V2=(153 atm)(13.2 L)(310 K)/(300 K)(3.13 atm)=667 L

An ideal gas is held in a rigid container at temperature of 201∘C. Upon cooling the gas, the pressure decreases by a factor of 2.0, while the volume and the amount of gas remain constant. What must be the final temperature of the gas?

≈−36∘C P1/T1=P2/T2 Notice that because the pressure decreases by a factor of two, we can write P2=1.0/2.0P1. We can substitute 1.0/2.0P1 for P2, thus cancelling P1. T2=474.15K/2.0=237.075K Convert back to Celsius and round to the nearest degree. 237.075K−273.15=−36.075∘C≈−36∘C

Why is it important to know gas properties at STP? Because comparison of properties is possible only if the properties are reported against a standard temperature and pressure. Gases can only react at STP. Gases must be stored at STP. In order to know that exactly one mole of an ideal gas has a volume of 22.4L.

-Because comparison of properties is possible only if the properties are reported against a standard temperature and pressure. -In order to know that exactly one mole of an ideal gas has a volume of 22.4L. Chemists sometimes make comparisons against a standard temperature and pressure (STP) for reporting properties of gases. At STP, one mole of an ideal gas has a volume of 22.4L. This is referred to as the standard molar volume.

Hydrogen sulfide gas (H2S) is a highly toxic gas that is responsible for the smell of rotten eggs. The volume of a container of hydrogen sulfide is 44.2mL. After the addition of more hydrogen sulfide, the volume increases to 98.5mL under constant pressure and temperature. The container now holds 1.97×10−3mol of the gas. How many grams of hydrogen sulfide were in the container initially? Give your answer in three significant figures.

0.0301g of hydrogen sulfide.

A container holds 3.58×10−3mol of the noble gas krypton. After the addition of 5.61×10−3mol of krypton, the volume of the container increases under constant temperature and pressure to 0.206L. What was the initial volume of the container?

0.0802L First, calculate the final number of moles of krypton (n2) in the container. n2=n1+nadded= n2= 3.58×10^−3mol + 5.61×10^−3mol n2=9.19×10−3mol Rearrange Avogadro's law to solve for V1.V1=V2×n1/n2Substitute in the known values of n1, n2, and V2.V1=0.206L×3.58×10−3mol/9.19×10−3mol=0.0802L

Sulfur hexafluoride is a non-toxic gas that is used to trace airflow in buildings. A 12.2L container of sulfur hexafluoride contains 0.547mol of gas. More sulfur hexafluoride is added to the container to increase its volume to 17.5L, while the temperature and pressure remain constant. How many additional moles of sulfur hexafluoride gas were added to the container?

0.238mol n2=V2×n1/V1 Substitute in the known values of n1, V1, and V2. n2=17.5L×0.547mol/12.2L=0.785mol Find the difference between the final number of moles (n2) and the initial number of moles (n1). n2−n1=0.785mol−0.547mol=0.238mol

Chlorine gas is held in a container with volume of 1.15L. The container initially held 3.53×10−2mol of chlorine gas, but after some of the gas was removed to carry out a reaction, the volume decreased to 0.790L while the temperature and pressure remained constant. How many grams of chlorine gas were removed from the container? The molar mass of chlorine gas (Cl2) is 70.9gmol. Give your answer in three significant figures.

0.783g of chlorine gas n2=V2×n1/V1 Substitute the known values for n1, V1, and V2. n2=0.790L×3.53×10^−2mol/1.15L=2.425×10^−2mol Find the difference between the initial number of moles (n1) and the final number of moles (n2). n1−n2=3.53×10^−2mol−2.425×10^−2mol=1.105×10^−2mol Finally, convert the number of moles lost to grams using the molar mass. mchlorine=1.105×10^−2mol×70.9g/mol=0.783g

0.23mol of a gas fills a volume of 0.550L. Additional gas was forced into the container under conditions of constant temperature and pressure, and as a result the container expanded to a volume of 2.20L. How many moles of gas are in the container after the expansion?

0.92mol of gas

0.23mol of a gas fills a volume of 0.550L. Additional gas was forced into the container under conditions of constant temperature and pressure, and as a result the container expanded to a volume of 2.20L. How many moles of gas are in the container after the expansion?

0.92mol of gas We know V1=0.550L, n1=0.23mol, and V2=2.20L. Use Avogadro's law to determine the number of moles after the expansion of the gas, n2. V1/n1=V2/n2 n2=V2×n1/V1 n2=2.20L×0.23mol/0.550L n2=0.92mol

Which of the following represents the conditions of standard temperature and pressure (STP)? 25∘C and 1kPa 0∘C and 101.325kPa 75∘F and 1atm 298.15K and 101.325atm 273.15K and 1atm

0∘C and 101.325kPa 273.15K and 1atm Standard conditions of temperature and pressure are defined as 273.15K(0∘C) and 1atm(101.325kPa).

Methane, CH4, is being considered for use as an alternative automotive fuel to replace gasoline. One gallon of gasoline could be replaced by 655 g of CH4. What is the volume of this much methane at 25∘C and 745 torr?

1.02×10^3 L n=655 g CH4×1 mol/16.043 g CH4=40.8 mol T=25∘C+273=298 K P=745torr×1 atm/760torr=0.980 atm V=nRT/P =(40.8 mol)(0.08206 L atm/mol K)(298 K)/0.980 atm=1.02×10^3 L