6.10: Gas Law Equations: Calculations

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Learning Objectives

Apply a Gas Law equation to calculate the value of an unknown that corresponds to one of the principal measurable quantities of gases.

The Gas Law relationships that have been discussed in the previous sections of this chapter are moderately challenging to utilize, as several variables are involved in each corresponding equation. The following paragraphs will describe and apply the process that should be used to solve a Gas Law problem.

The numerical quantities that are incorporated into Gas Law equations are often presented within the context of a word problem. Unlike the molar quantities that were developed in Chapter 4 and the heat transfer equations that were discussed in Chapter 5, the Gas Laws equations do not have corresponding indicator words or phrases. Instead, in order to select the Gas Law that should be applied to solve a problem, each numerical quantity that is given in the problem should be assigned to a variable. Because each Gas Law equation relates a unique combination of pressure, volume, temperature, and amount-based quantities, only a single Gas Law will correspond to the values that are given in a particular problem.

Before the selected Gas Law equation can be applied, the validity of the units that are associated with the given numerical values must be confirmed. As stated in Section 6.2, the pressure of a gas must be expressed in atmospheres (atm), millimeters of mercury (mmHg), torr (Torr), or pascals (Pa). The volume of a gas must be given in liters (L) or milliliters (mL). The temperature and the amount of gas that is present must be reported in Kelvin (K) and moles (mol), respectively. Furthermore, recall that a Gas Law equation contains both an "initial" and a "final" variable for each of the quantities that it relates. Therefore, in order to achieve unit cancelation through multiplication or division, which are the mathematical operations that are utilized in these equations, the units in which the initial and final values are expressed must be identical for a given variable.

A variety of algebraic processes can be applied to isolate the variable that corresponds to the unknown quantity. Because, with the exception of Boyle's Law, each Gas Law contains both a numerator and a denominator, the mathematical process of cross-multiplication and division is commonly utilized to solve these equations. When cross-multiplying, the numerator on one side of an equation is multiplied by the denominator on the other side of the equal sign. The process is repeated for the remaining quantities, and the mathematical products are equated to one another. Finally, both sides of the resultant equation must be divided by the values of any quantities that are multiplicatively-related to the unknown quantity, in order to isolate its corresponding variable. The unit that remains uncanceled becomes the unit on the final answer, which is calculated by multiplying any values that are present in the numerator of the resultant fraction and dividing by the quantities in its denominator.

For example, consider a balloon that is filled with 3.4 liters of helium gas at 55.0 degrees Celsius. Calculate the volume of the balloon at -12.8 degrees Celsius, if the pressure and amount of gas in the balloon are held constant.

In order to select the Gas Law that should be applied to solve this problem, each numerical quantity that is given in the problem must be assigned to a variable. Additionally, in order to be incorporated into the selected equation, the validity of the units that are associated with the given numerical values must be confirmed. As stated above, the pressure of a gas must be expressed in atmospheres (atm), millimeters of mercury (mmHg), torr (Torr), or pascals (Pa), the volume of a gas must be given in liters (L) or milliliters (mL), and the temperature and the amount of gas that is present must be reported in Kelvin (K) and moles (mol), respectively. Furthermore, in order to achieve unit cancelation through multiplication or division, which are the mathematical operations that are utilized in the Gas Law equations, the units in which the initial and final values are expressed must be identical for a given variable.

The numerical values that are given in the problem, the variables to which these quantities are assigned, and an indication of the validity of their corresponding units are shown in the following table.

Numerical Quantity	Variable	Unit Validity
3.4 L	V₁
55.0 °C	T₁
V₂	V₂
-12.8 °C	T₂

Because the values that are given in the problem correspond to volume and temperature measurements, Charles's Law must be applied to solve the given problem. Additionally, since V₂ is the only variable that cannot be assigned to a numerical value in the given problem, the final volume of the gas is the unknown quantity that will be calculated upon solving this Gas Law equation. The problem does not specify whether the final answer should be expressed in liters or milliliters. Therefore, the given unit for the initial volume, "liters," is acceptable for this problem. Of the remaining variables, neither temperature is reported in a valid unit. Therefore, as shown below, both the initial temperature, T₁,

T_K = 55.0 °C + 273.15
T_K = 328.15 K ≈ 328.2 K

and the final temperature, T₂,

T_K = -12.8 °C + 273.15
T_K = 260.35 K ≈ 260.4 K

must both be converted to Kelvin before they can be incorporated into the Charles's Law equation.

The updated numerical values that are summarized in the following table are all expressed in the appropriate units and, therefore, can be utilized to solve the given problem.

Numerical Quantity	Variable	Unit Validity
3.4 L	V₁
328.2 K	T₁
V₂	V₂
260.4 K	T₂

The quantities that are shown in the table above can now be incorporated into the Charles's Law equation, as shown below. In order to solve for V₂, the quantities in this equation can be cross-multiplied. During the subsequent division step, the temperature unit is canceled, because "K" is present in the numerator and the denominator in the resultant fraction. The unit that remains after this cancelation is "L," which is a valid unit for expressing the volume of a gas. The numerical solution is calculated by multiplying the values that are present in the numerator of this fraction and then dividing by the quantities in its denominator. When using a calculator, the "=" key should be used after each division. Applying the correct number of significant figures to the calculated quantity results in the final answer that is shown below.

\( \dfrac{ \rm{V_{1}}}{\rm{T_{1}}} \) = \( \dfrac{ \rm{V_{2}}}{\rm{T_{2}}} \)

\( \dfrac{3.4 \; \rm{L}}{328.2 \; \rm{K}} \) = \( \dfrac{ \rm{V_{2}}}{260.4 \; \rm{K}} \)

(\({3.4 \; \rm{L}}\))(\({260.4 \; \rm{K}}\)) = (\(\rm{V_{2}}\))(\({328.2 \; \rm{K}}\))

\(\rm{V_{2}}\) = \({\dfrac{({3.4 \; \rm{L}}) ({260.4 \; \cancel{\rm{K}}})} {({328.2 \; \cancel{\rm{K}}})}}\)

\(\rm{V_{2}}\) = \({2.697623... \; \rm{L}} ≈ {2.7 \; \rm{L}}\)

Finally, recall that the quantities that are related by Charles's Law are directly proportional. Therefore, since the temperature of the gas decreased, its volume must also decrease. Because the calculated final volume, V₂, is smaller than the initial volume, V₁, the value of the final answer is reasonable.

Exercise \(\PageIndex{1}\)

Consider a 0.41 liter balloon that is pressurized with 1,260 millimeters of mercury of water vapor at 595 Kelvin. Calculate the pressure of the gas if its volume is quadrupled at 307 Kelvin and the amount of gas in the balloon is held constant.

Answer

Numerical Quantity	Variable	Unit Validity
1,260 mmHg	P₁
0.41 L	V₁
595 K	T₁
P₂	P₂
"Quadrupled"	V₂
307 K	T₂

Because the values that are given in the problem correspond to pressure, volume, and temperature measurements, the Combined Gas Law must be applied to solve the given problem. Additionally, since P₂ is the only variable that cannot be assigned to a numerical value in the given problem, the final pressure of the gas is the unknown quantity that will be calculated upon solving this Gas Law equation. The problem does not specify whether the final answer should be expressed in atmospheres, millimeters of mercury, torr, or pascals. Therefore, the given unit for the initial pressure, "millimeters of mercury," is acceptable for this problem. The numerical values that are summarized in the table that is shown above are all expressed in the appropriate units and, therefore, can be utilized to solve the given problem. However, the final volume of the gas is not given as a numerical quantity. Instead, the problem indicates that the volume of the gas "quadrupled." Therefore, in order to calculate the final volume of the gas, the initial volume must be multiplied by four, as shown below.

V₂ = 4(0.41 L)
V₂ = 1.64 L ≈ 1.6 L

The updated numerical values that are summarized in the following table are all expressed in the appropriate units and, therefore, can be utilized to solve the given problem.

Numerical Quantity	Variable	Unit Validity
1,260 mmHg	P₁
0.41 L	V₁
595 K	T₁
P₂	P₂
1.6 L	V₂
307 K	T₂

The quantities that are shown in the table above can now be incorporated into the Combined Gas Law equation, as shown below. In order to solve for P₂, the quantities in this equation can be cross-multiplied. During the subsequent division step, the volume and temperature units, "L" and "K," respectively, are canceled, because both of these units are present in the numerator and the denominator in the resultant fraction. The unit that remains after these cancelations is "mmHg," which is a valid unit for expressing the pressure of a gas. The numerical solution is calculated by multiplying the values that are present in the numerator of this fraction and then dividing by the quantities in its denominator. When using a calculator, the "=" key should be used after each division. Applying the correct number of significant figures to the calculated quantity results in the final answer that is shown below.

\( \dfrac{\rm{P_{1}} \rm{V_{1}}}{\rm{T_{1}}} \) = \( \dfrac{\rm{P_{2}} \rm{V_{2}}}{\rm{T_{2}}} \)

\({\dfrac{({1,260 \; \rm{mmHg}}) ({0.41 \; \rm{L}})} {({595 \; \rm{K}})}}\) = \({\dfrac{({\rm{P_{2}}}) ({1.6 \; \rm{L}})} {({307 \; \rm{K}})}}\)

(\({1,260 \; \rm{mmHg}}\))(\({0.41 \; \rm{L}}\))(\({307 \; \rm{K}}\)) = (\(\rm{P_{2}}\))(\({1.6 \; \rm{L}}\))(\({595 \; \rm{K}}\))

\(\rm{P_{2}}\) = \({\dfrac{({1,260 \; \rm{mmHg}}) ({0.41 \; \cancel{\rm{L}}}) ({307 \; \bcancel{\rm{K}}})} {({1.6 \; \cancel{\rm{L}}}) ({595 \; \bcancel{\rm{K}}})}}\)

\(\rm{P_{2}}\) = \({166.5926... \; \rm{mmHg}} ≈ {170 \; \rm{mmHg}}\)

Finally, recall that since three variables are incorporated into the Combined Gas Law, its collective quantities are neither directly nor indirectly proportional. Therefore, the validity of the final answer cannot be confirmed by comparing its value to the initial pressure that was given in the problem.

Exercise \(\PageIndex{2}\)

Consider a tank that is pressurized with 913 torr of molecular oxygen gas at 599 degrees Fahrenheit. Calculate the temperature of the gas if 548 torr of molecular oxygen gas is added and the volume and amount of gas in the balloon are held constant.

Answer

Numerical Quantity	Variable	Unit Validity
913 torr	P₁
599 °F	T₁
"Added"	P₂
T₂	T₂

Because the values that are given in the problem correspond to pressure and temperature measurements, Gay-Lussac's Law must be applied to solve the given problem. Additionally, since T₂ is the only variable that cannot be assigned to a numerical value in the given problem, the final temperature of the gas is the unknown quantity that will be calculated upon solving this Gas Law equation. While the problem does not specify a unit for the final answer, any temperature values that are incorporated into Gas Law equations must be expressed in Kelvin, as stated above. Therefore, the initial temperature, T₁, is not expressed in a valid unit and must be converted to Kelvin, as shown below, before it can be incorporated into the Gay-Lussac's Gas Law equation.

599 °F = 1.8T_C + 32
567 ° = 1.8T_C
T_C = 315 °C

T_K = 315 °C + 273.15
T_K = 588.15 K ≈ 588 K

Additionally, the final pressure of the gas is not explicitly stated. Instead, the problem indicates that "548 torr of molecular oxygen gas is added" to the tank. Therefore, in order to calculate the final pressure of the gas, this quantity must be added to the initial pressure of the gas, as shown below.

P₂ = 913 torr + 548 torr
P₂ = 1,461 torr

The updated numerical values that are summarized in the following table are all expressed in the appropriate units and, therefore, can be utilized to solve the given problem.

Numerical Quantity	Variable	Unit Validity
913 torr	P₁
588 K	T₁
1,461 torr	P₂
T₂	T₂

The quantities that are shown in the table above can now be incorporated into the Gay-Lussac's Gas Law equation, as shown below. In order to solve for T₂, the quantities in this equation can be cross-multiplied. During the subsequent division step, the pressure unit is canceled, because "torr" is present in the numerator and the denominator in the resultant fraction. As any temperature value that is calculated using a Gas Law equation must be expressed in Kelvin, as stated above, the unit that remains after this cancelation, "K," is appropriate for reporting the final answer. The numerical solution is calculated by multiplying the values that are present in the numerator of this fraction and then dividing by the quantities in its denominator. When using a calculator, the "=" key should be used after each division. Applying the correct number of significant figures to the calculated quantity results in the final answer that is shown below.

\( \dfrac{\rm{P_{1}}}{\rm{T_{1}}} \) = \( \dfrac{\rm{P_{2}}}{\rm{T_{2}}} \)

\({\dfrac{913 \; \rm{torr}} {588 \; \rm{K}}}\) = \({\dfrac{1,461 \; \rm{torr}} {\rm{T_{2}}}}\)

(\({913 \; \rm{torr}}\))(\({\rm{T_{2}}}\)) = (\({1,461 \; \rm{torr}}\))(\({588 \; \rm{K}}\))

\(\rm{T_{2}}\) = \({\dfrac{({1,461 \; \cancel{\rm{torr}}}) ({588 \; \rm{K}})} {({913 \; \cancel{\rm{torr}}})}}\)

\(\rm{T_{2}}\) = \({940.9288... \; \rm{K}} ≈ {941 \; \rm{K}}\)

Finally, recall that the quantities that are related by Gay-Lussac's Law are directly proportional. Therefore, since the pressure of the gas increased, its temperature must also increase. Because the calculated final temperature, T₂, is larger than the initial temperature, T₁, the value of the final answer is reasonable.

Exercise \(\PageIndex{3}\)

Consider a 575 milliliter balloon that is pressurized with 701,000 pascals of carbon dioxide gas. Calculate the volume of the balloon, in liters, if its pressure is changed to 3.3 atmospheres and the temperature and amount of gas in the balloon are held constant.

Answer

Numerical Quantity	Variable	Unit Validity
701,000 Pa	P₁
575 mL	V₁
3.3 atm	P₂
V₂	V₂

Because the values that are given in the problem correspond to pressure and volume measurements, Boyle's Law must be applied to solve the given problem. Additionally, since V₂ is the only variable that cannot be assigned to a numerical value in the given problem, the final volume of the gas is the unknown quantity that will be calculated upon solving this Gas Law equation. The problem specifies that the final answer must be expressed in liters, which is not consistent with the given unit for the initial volume, "milliliters." In order to remedy this discrepancy, milliliters must be converted to liters. Because milliliters is a valid unit for expressing the volume of a gas, incorporating the given value into the Boyle's Law equation is acceptable, and the calculated answer would then need to be converted to liters. Alternatively, the initial volume could be immediately converted to liters, as shown below. As liters is also a valid unit for reporting the volume of a gas, this value can be used in the Boyle's Law equation.

\( {\text {575}} {\cancel{\rm{mL} }} \times\) \( \dfrac{\rm{L} }{1,000 \; \cancel{\rm{mL} }}\) = \( {\text {0.575}} \; \rm{L} \)

Additionally, the units that are associated with the given values for the initial pressure, P₁, and the final pressure, P₂, of the gas are not consistent with one another and, therefore, will not cancel when incorporated into the Boyle's Law equation. In order to remedy this discrepancy, one of these units must be converted to match the other. Because both of the given units, "pascals" and "atmospheres," are valid units for expressing the pressure of a gas, altering either unit is acceptable. The conversion of P₁ to a corresponding quantity in atmospheres is shown below.

\( {\text {701,000}} {\cancel{\rm{Pa} }} \times\) \( \dfrac{1 \; \rm{atm} }{101,325 \; \cancel{\rm{Pa} }}\) = \( {\text {6.918332...}} \; \rm{atm} \) ≈ \( {\text {6.91}} \; \rm{atm} \)

The updated numerical values that are summarized in the following table are all expressed in the appropriate units and, therefore, can be utilized to solve the given problem.

Numerical Quantity	Variable	Unit Validity
6.91 atm	P₁
0.575 L	V₁
3.3 atm	P₂
V₂	V₂

The quantities that are shown in the table above can now be incorporated into the Boyle's Law equation, as shown below. In order to solve for V₂, the quantities on both sides of this equation must be divided by the value of P₂, "3.3 atm," which causes the cancelation of the pressure unit, "atm," as this unit is present in the numerator and the denominator in the resultant fraction. The unit that remains after this cancelation is "liters," which, per the information in the given problem, is the unit in which the final volume of the gas must be expressed. The numerical solution is calculated by multiplying the values that are present in the numerator of this fraction and then dividing by the quantities in its denominator. When using a calculator, the "=" key should be used after each division. Applying the correct number of significant figures to the calculated quantity results in the final answer that is shown below.

\({\rm{P_{1}}}\)\({\rm{V_{1}}}\) = \(\rm{P_{2}}\)\({\rm{V_{2}}}\)

(\({6.91 \; \rm{atm}}\))(\({0.575 \; \rm{L}}\)) = (\(3.3 \; \rm{atm}\))(\({\rm{V_{2}}}\))

\(\rm{V_{2}}\) = \({\dfrac{({6.91 \; \cancel{\rm{atm}}}) ({0.575 \; \rm{L}})} {({3.3 \; \cancel{\rm{atm}}})}}\)

\(\rm{V_{2}}\) = \({1.204015... \; \rm{L}} ≈ {1.2 \; \rm{L}}\)

Finally, recall that the quantities that are related by Boyle's Law are indirectly proportional. Therefore, since the pressure of the gas decreased, its volume must increase. Because the calculated final volume, V₂, is larger than the initial volume, V₁, the value of the final answer is reasonable.

Exercise \(\PageIndex{4}\)

Consider a 125 milliliter balloon that contains 59.76 grams of argon gas. Calculate the number of atoms of argon that are contained in the balloon, if its volume is expanded to 915 milliliters and the pressure and temperature of the gas are held constant.

Answer

Numerical Quantity	Variable	Unit Validity
125 mL	V₁
59.76 g Ar	n₁
915 mL	V₂
n₂	n₂

Because the values that are given in the problem correspond to volume and amount-based measurements, Avogadro's Law must be applied to solve the given problem. Additionally, since n₂ is the only variable that cannot be assigned to a numerical value in the given problem, the final amount of gas is the unknown quantity that will be calculated upon solving this Gas Law equation. The problem specifies that the final amount of gas must be expressed in atoms of argon, which is not consistent with the given unit for the initial amount of gas that is present, "grams." Furthermore, any amount-based quantities that are incorporated into Gas Law equations must be expressed in moles, as stated above. Therefore, the initial amount of gas, n₁, is not expressed in a valid unit and must be converted to moles before it can be incorporated into the Avogadro's Law equation. In order to convert the given amount of gas to moles, a molar equality must be developed and applied. In particular, the mass unit "grams" indicates that a mass-based equality is required for this conversion. Since argon, Ar, is an element, an atomic weight equality for this chemical should be developed by equating the atomic weight of the element, which should be reported to the hundredths place, to 1 mol of the element, and utilizing an elemental symbol as the secondary unit on both sides of the resultant equality, as shown below.

1 mol Ar = 39.95 g Ar

In order to eliminate the given unit for the initial amount of gas that is present, "grams of argon," the equality that is written above must be applied as a conversion factor. Applying the correct number of significant figures to the calculated quantity results in the numerical value that is shown below.

\( {\text {59.76}}\) \({\cancel{\rm{g} \; \rm{Ar}}} \times\) \( \dfrac{1 \; \rm{mol} \; \rm{Ar}}{39.95\; \cancel{\rm{g} \; \rm{Ar}}}\) = \( {\text {1.4958698...}}\) \({\rm{mol} \; \rm{Ar}}\) ≈ \( {\text {1.496}}\) \({\rm{mol} \; \rm{Ar}}\)

The updated numerical values that are summarized in the following table are all expressed in the appropriate units and, therefore, can be utilized to solve the given problem.

Numerical Quantity	Variable	Unit Validity
125 mL	V₁
1.496 mol Ar	n₁
915 mL	V₂
n₂	n₂

The quantities that are shown in the table above can now be incorporated into the Avogadro's Law equation, as shown below. In order to solve for n₂, the quantities in this equation can be cross-multiplied. During the subsequent division step, the volume unit is canceled, because "mL" is present in the numerator and the denominator in the resultant fraction. As any amount-based quantity that is calculated using a Gas Law equation must be expressed in moles, as stated above, the unit that remains after this cancelation, "mol," is appropriate for reporting the final amount of gas that is present in the balloon. The numerical solution is calculated by multiplying the values that are present in the numerator of this fraction and then dividing by the quantities in its denominator. When using a calculator, the "=" key should be used after each division. Applying the correct number of significant figures to the calculated quantity results in the answer that is shown below.

\( \dfrac{\rm{V_{1}}}{\rm{n_{1}}} \) = \( \dfrac{\rm{V_{2}}}{\rm{n_{2}}} \)

\({\dfrac{125 \; \rm{mL}} {1.496 \; \rm{mol} \; \rm{Ar}}}\) = \({\dfrac{915 \; \rm{mL}} {\rm{n_{2}}}}\)

(\({125 \; \rm{mL}}\))(\({\rm{n_{2}}}\)) = (\({915 \; \rm{mL}}\))(\({1.496 \; \rm{mol} \; \rm{Ar}}\))

\(\rm{n_{2}}\) = \({\dfrac{({915 \; \cancel{\rm{mL}}}) ({1.496 \; \rm{mol} \; \rm{Ar}})} {({125 \; \cancel{\rm{mL}}})}}\)

\(\rm{n_{2}}\) = \({10.95072 \; \rm{mol} \; \rm{Ar}} ≈ {11.0 \; \rm{mol} \; \rm{Ar}}\)

Recall that the quantities that are related by Avogadro's Law are directly proportional. Therefore, since the volume of the gas increased, the amount of gas must also increase. Because the calculated final amount of gas, n₂, is larger than the initial amount of gas, n₁, the numerical value of the solution that is shown above is reasonable.

However, the problem specifies that the final amount of gas must be expressed in atoms of argon, which is not consistent with the unit that is shown above. In order to convert the final amount of gas to atoms of argon, a molar equality must be developed and applied. In particular, the word "atoms" indicates that an Avogadro's number equality is required for this conversion. Since "atoms" is an indicator word, this word is inserted as the second unit on the right side of this equality, and the chemical formula for argon, Ar, is incorporated into both of the secondary unit positions, as shown below. Because the given chemical information is the name of an element, the indicator word "atoms" appropriately corresponds to the chemical formula that is applied in this equality.

1 mol Ar = 6.02 × 10²³ Ar atoms

In order to completely eliminate the unit "mol Ar," the equality that is written above must be utilized as a conversion factor. Applying the correct number of significant figures to the calculated quantity results in the final answer that is shown below.

\( {11.0 \; \cancel{\rm{mol} \; \rm{Ar}}} \times\) \( \dfrac{6.02 \times 10^{23} \; \rm{atoms} \; \rm{Ar}}{1 \; \cancel{\rm{mol} \; \rm{Ar}}}\) = \( {\text {6.622}} \times {\text{10}^{24}}\) \({\rm{atoms} \; \rm{Ar}}\) ≈ \( {\text {6.62}} \times {\text{10}^{24}}\) \({\rm{atoms} \; \rm{Ar}}\)