What steps should be followed when using the ideal gas law to solve a problem?

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Which of the post-obit is a condition of the ideal gas constabulary?

Possible Answers:

The gas molecules must interact with each other

The gas molecules must adhere to the container wall

The gas molecules must be in specific mole ratios

The gas molecules do non have book

The gas molecules must be moving at a tedious speed

Correct answer:

The gas molecules do not accept volume

Explanation:

The ideal gas law has some conditions that must exist met, atmospheric condition that certainly cannot be met in the real world. These conditions include that the gases cannot collaborate with ane another, gases must be moving in a random direct-line fashion, gas molecules must non accept upwardly any space, and gases must be in perfect elastic collisions with the walls of the container. These atmospheric condition minimize the effect that gas molecules have on one other, allowing a prediction based on completely random and unimpeded molecular motion. In reality, these conditions are impossible. All real gas molecules have a molecular volume and some degree of intermolecular attraction forces.

A $\small 4.5L$ container of gas has a pressure level of $\small 3.0atm$ at a temperature of $\small 100^oC$ . The container is expanded to $\small 6L$ , and the temperature is increased to $\small 200^oC$ .

What is the final pressure of the container?

Correct answer:

2.9 atm

Explanation:

In this instance, two variables are changed between the initial and final containers: volume and temperature. Since we are looking for the concluding pressure on the container, we tin use the combined gas law in guild to solve for the final pressure:

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$

When using the ideal gas law, remember that temperature must be in Kelvin, not Celsius, so nosotros will need to convert.

T_1=100^oC+273=373K

T_2=200^oC+273=473K

Apply the given values to solve for the final pressure.

$\frac{(3atm)(4.5L)}{373K} = \frac{(P_{2})(6L)}{473K}$

$P_{2} = 2.9atm$

At STP, an unknown gas has a density of $\small 1.7\frac{g}{L}$ .

Based on this information and the periodic table, what is the identity of the gas in the container?

$R=0.08206\frac{L*atm}{mol*K}$

Possible Answers:

Krypton

Nitrogen gas

Neon

Fluorine gas

Right answer:

Fluorine gas

Explanation:

Once once more, nosotros can use the ideal gas police in guild to solve for the unknown gas. Nosotros will start again with the ideal gas constabulary:

PV = nRT

Density has the units of mass over book, which ways that we tin rearrange the platonic gas law then that density is on ane side of the equation. Nonetheless, nosotros demand to substitute i of the values so mass is also in the equation. Remember that moles are equal to mass divided by molar mass, so we can rewrite "n" in guild to solve for density:

$PV = \frac{m}{MM}RT$

$\frac{m}{V} = \frac{PMM}{RT}$

STP means that the container is at i atmosphere of pressure and 273 Kelvin. Knowing this, we can solve for molar mass:

$1.7\frac{g}{L} = \frac{(1atm)(MM)}{(.08206)(273K)}$

$MM = 38.1 \frac{g}{mol}$

Fluorine gas has a molar mass of 38.00 grams per mol. As a consequence, we decide that fluorine gas is the unknown gas in the container.

$\small 10g$ of an unknown gas are independent in a $\small 3L$ container. The container has a pressure of $\small 2.04atm$ at a temperature of $\small 25^oC$ .

Based on this information and the periodic table, which of the following gases is in the container?

$R=0.08206\frac{L*atm}{K*mol}$

Possible Answers:

Neon

Nitrogen gas

Argon

Helium

Explanation:

In order to determine the gas being contained, we are going to need to rewrite the ideal gas law. The platonic gas law is written as follows:

PV = nRT

The number of moles of a gas can be rewritten every bit the mass of the gas divided by its molar mass. Knowing this, we can rewrite the equation, and solve for the tooth mass of the gas.

$n=\frac{m}{MM}$

$PV = (\frac{m}{MM})RT$

$MM = \frac{mRT}{PV}$

Utilise the given values in this equation to solve for the molar mass. Call up to first convert degree Celsius to Kelvin!

T=25^oC+273=298K

$MM = \frac{(10g)(0.08206\frac{L*atm}{mol*K})(298K)}{(2.04atm)(3L)} = 39.95\frac{g}{mol}$

At present that we accept solved for the molar mass, we can see which gas has this molar mass on the periodic tabular array. Argon has a molar mass of 39.95 grams per mole, so nosotros can determine that argon is the gas in the container.

Which police force is represented by the following formula?

$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}$

Possible Answers:

Dalton'south law

Combined gas law

Charles'southward police

Boyle's law

Platonic gas law

Correct respond:

Combined gas law

Explanation:

The combined gas police force takes Boyle'southward, Charles's, and Gay-Lussac's police force and combines it into one law.

$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}$

It is able to chronicle temperature, pressure, and volume of i system when the parameters for any of the three change.

Boyle's law relates force per unit area and volume: P_1V_1=P_2V_2

Charles'southward police relates temperature and volume: $\frac{V_1}{T_1}=\frac{V_2}{T_2}$

Gay-Lussac's constabulary relates temperature and pressure: $\frac{P_1}{T_1}=\frac{P_2}{T_2}$

The platonic gas law relates temperature, pressure, volume, and moles in coordination with the ideal gas constant: PV=nRT

A scuba diver uses compressed air to breath nether water. He starts with an air volume of 3.20L at bounding main level ( 1atm ) at a temperature of 30^oC . What is the volume of air in his tank at a depth of 100ft ( 3atm ) and a temperature of 24^oC ?

Correct reply:

1.05L

Explanation:

This question requires the combined equation of the individual gas laws:

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

To use this equation, we offset need to convert the given temperatures to Kelvin.

30^oC+273=303K

24^oC+273=297K

We now know the initial pressure level, volume, and temperature, assuasive us to solve the left side of the equation.

$\frac{(1atm)(3.20L)}{303K}=\frac{P_2V_2}{T_2}$

$0.0106\frac{atm\cdot L}{K}=\frac{P_2V_2}{T_2}$

Use the given values for the final temperature and force per unit area to solve for the final volume.

$0.0106\frac{atm\cdot L}{K}=\frac{(3atm)V_2}{297K}$

$0.0106\frac{atm\cdot L}{K}=(0.0101\frac{atm}{K})V_2$

$V_2=\frac{0.0106\frac{atm\cdot L}{K}}{0.0101\frac{atm}{K}}$

V_2=1.05L

A 400mL sample of helium gas at 293K and a force per unit area of 2.2atm is used to inflate a airship. What is the book this gas occupies when the temperature reaches 308K at a pressure of 1.4atm during the aggrandizement?

Right reply:

661mL

Explanation:

This question requires s to employ the combined gas law:

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

We know the initial pressure, temperature, and book, allowing united states of america to solve for the left side of the equation.

$\frac{(2.2atm)(400mL)}{292K}=\frac{P_2V_2}{T_2}$

$3.00\frac{atm\cdot mL}{K}=\frac{P_2V_2}{T_2}$

We are likewise given the final pressure and temperature. Using these values on the right side of the equation nosotros can solve for the final volume.

$3.00\frac{atm\cdot mL}{K}=\frac{(1.4atm)V_2}{308K}$

$3.00\frac{atm\cdot mL}{K}=(0.00455\frac{atm}{K})V_2$

$V_2=\frac{3.00\frac{atm\cdot mL}{K}}{0.00455\frac{atm}{K}}$

V_2=661mL

Below is the ideal gas law.

PV=nRT

What does each letter of the alphabet of the ideal gas law represent?

Possible Answers:

P: pressure
5: book
n: moles
R: charge per unit
T: time

None of these

P: potential energy
5: velocity
n: moles
R: ideal gas constant
T: total free energy

P: pressure
V: book
north: moles
R: platonic gas constant
T: temperature

P: pressure
Five: volume
n: moles
R: ideal gas constant
T: time

Correct answer:

P: pressure
V: volume
due north: moles
R: platonic gas constant
T: temperature

Explanation:

The ideal gas law is used to identify values in a given state (for the values of pressure, volume, number of moles, and temperature) for an platonic, hypothetical gas. Because no gases are truly ideal, this only works as an approximation, and some gases are more than ideal than others.

A gas in a $450\,mL$ container is at STP. How many moles of gas are in the container?

Right respond:

$0.020\, moles$

Explanation:

First we need to remember what STP ways since it'southward important. This is "standard temperature and pressure". Because we are looking for the number of moles nosotros need the ideal gas constabulary.

PV=nRT

At STP we will take the temperature to be 273K and the force per unit area to exist $1\,atm$ . Solving for the number of moles gives

$n=\frac{PV}{RT}=\frac{(1\,atm)(0.450L)}{(0.0821\frac{atm\cdot L}{mol\cdot K})(273K)}$

$n=0.020\, moles$

Notice that the units of the gas constant R hateful we must accept a volume in liters. The volume was given in mL and therefore had to be converted.

A gas at STP has a book of $50.0\,mL$ . The gas is compressed to $22.1\,mL$ and heated to 400K . What is the new pressure of the gas in atm?

Correct respond:

$P_{2}=3.31\,atm$

Caption:

We have a lot going on in this problem. Since both temperature and book change and we want to know the final pressure, the combined gas constabulary will be used:

$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$

Since the gas is at STP to showtime we know the pressure ( $1\,atm$ ) and the temperature ( 273K ). Standard pressure has multiple values depending on the units, but we desire pressure in atm for our final respond so we will choose the advisable value and units for STP. Solving for the terminal pressure gives:

$P_{2}=\frac{P_{1}V_{1}T_{2}}{T_{1}V_{2}}$

Plugging in everything gives:

$P_{2}=\frac{(1\,atm)(50\,mL)(400K)}{(273K)(22.1\,mL)}=3.31\,atm$

We expect that raising the temperature volition increase the pressure as well as reducing the volume. Since both actions will increase the pressure we must accept a final pressure level greater than the initial pressure.

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What steps should be followed when using the ideal gas law to solve a problem?

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All High School Chemistry Resources

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