6.10: Gas Law Equations: Calculations
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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 | V_{1} | |
55.0 °C | T_{1} | |
V_{2} | V_{2} | |
-12.8 °C | T_{2} |
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 V_{2} 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, T_{1},
T_{K} = 55.0 °C + 273.15
T_{K} = 328.15 K ≈ 328.2 K
and the final temperature, T_{2},
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 | V_{1} | |
328.2 K | T_{1} | |
V_{2} | V_{2} | |
260.4 K | T_{2} |
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 V_{2}, 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, V_{2}, is smaller than the initial volume, V_{1}, the value of the final answer is reasonable.
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
- 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 1,260 mmHg P_{1} 0.41 L V_{1} 595 K T_{1} P_{2} P_{2} "Quadrupled" V_{2} 307 K T_{2} V_{2} = 4(0.41 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.
V_{2} = 1.64 L ≈ 1.6 L
Numerical Quantity Variable Unit Validity 1,260 mmHg P_{1} 0.41 L V_{1} 595 K T_{1} P_{2} P_{2} 1.6 L V_{2} 307 K T_{2} \( \dfrac{\rm{P_{1}} \rm{V_{1}}}{\rm{T_{1}}} \) = \( \dfrac{\rm{P_{2}} \rm{V_{2}}}{\rm{T_{2}}} \)
\({\dfrac{({1,260 \; \rm{mmHg}}) ({0.41 \; \rm{L}})} {({595 \; \rm{K}})}}\) = \({\dfrac{({\rm{P_{2}}}) ({1.6 \; \rm{L}})} {({307 \; \rm{K}})}}\)
(\({1,260 \; \rm{mmHg}}\))(\({0.41 \; \rm{L}}\))(\({307 \; \rm{K}}\)) = (\(\rm{P_{2}}\))(\({1.6 \; \rm{L}}\))(\({595 \; \rm{K}}\))
\(\rm{P_{2}}\) = \({\dfrac{({1,260 \; \rm{mmHg}}) ({0.41 \; \cancel{\rm{L}}}) ({307 \; \bcancel{\rm{K}}})} {({1.6 \; \cancel{\rm{L}}}) ({595 \; \bcancel{\rm{K}}})}}\)
\(\rm{P_{2}}\) = \({166.5926... \; \rm{mmHg}} ≈ {170 \; \rm{mmHg}}\)
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.
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
- 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 913 torr P_{1} 599 °F T_{1} "Added" P_{2} T_{2} T_{2} 599 °F = 1.8T_{C} + 32
567 ° = 1.8T_{C}
T_{C} = 315 °CT_{K} = 315 °C + 273.15
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.
T_{K} = 588.15 K ≈ 588 KP_{2} = 913 torr + 548 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.
P_{2} = 1,461 torr
Numerical Quantity Variable Unit Validity 913 torr P_{1} 588 K T_{1} 1,461 torr P_{2} T_{2} T_{2} \( \dfrac{\rm{P_{1}}}{\rm{T_{1}}} \) = \( \dfrac{\rm{P_{2}}}{\rm{T_{2}}} \)
\({\dfrac{913 \; \rm{torr}} {588 \; \rm{K}}}\) = \({\dfrac{1,461 \; \rm{torr}} {\rm{T_{2}}}}\)
(\({913 \; \rm{torr}}\))(\({\rm{T_{2}}}\)) = (\({1,461 \; \rm{torr}}\))(\({588 \; \rm{K}}\))
\(\rm{T_{2}}\) = \({\dfrac{({1,461 \; \cancel{\rm{torr}}}) ({588 \; \rm{K}})} {({913 \; \cancel{\rm{torr}}})}}\)
\(\rm{T_{2}}\) = \({940.9288... \; \rm{K}} ≈ {941 \; \rm{K}}\)
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, T_{2}, is larger than the initial temperature, T_{1}, the value of the final answer is reasonable.
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
- 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 701,000 Pa P_{1} 575 mL V_{1} 3.3 atm P_{2} V_{2} V_{2} \( {\text {575}} {\cancel{\rm{mL} }} \times\) \( \dfrac{\rm{L} }{1,000 \; \cancel{\rm{mL} }}\) = \( {\text {0.575}} \; \rm{L} \)
Additionally, the units that are associated with the given values for the initial pressure, P_{1}, and the final pressure, P_{2}, 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 P_{1} 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 P_{1} 0.575 L V_{1} 3.3 atm P_{2} V_{2} V_{2} \({\rm{P_{1}}}\)\({\rm{V_{1}}}\) = \(\rm{P_{2}}\)\({\rm{V_{2}}}\)
(\({6.91 \; \rm{atm}}\))(\({0.575 \; \rm{L}}\)) = (\(3.3 \; \rm{atm}\))(\({\rm{V_{2}}}\))
\(\rm{V_{2}}\) = \({\dfrac{({6.91 \; \cancel{\rm{atm}}}) ({0.575 \; \rm{L}})} {({3.3 \; \cancel{\rm{atm}}})}}\)
\(\rm{V_{2}}\) = \({1.204015... \; \rm{L}} ≈ {1.2 \; \rm{L}}\)
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, V_{2}, is larger than the initial volume, V_{1}, the value of the final answer is reasonable.
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
- 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 125 mL V_{1} 59.76 g Ar n_{1} 915 mL V_{2} n_{2} n_{2} 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 V_{1} 1.496 mol Ar n_{1} 915 mL V_{2} n_{2} n_{2} \( \dfrac{\rm{V_{1}}}{\rm{n_{1}}} \) = \( \dfrac{\rm{V_{2}}}{\rm{n_{2}}} \)
\({\dfrac{125 \; \rm{mL}} {1.496 \; \rm{mol} \; \rm{Ar}}}\) = \({\dfrac{915 \; \rm{mL}} {\rm{n_{2}}}}\)
(\({125 \; \rm{mL}}\))(\({\rm{n_{2}}}\)) = (\({915 \; \rm{mL}}\))(\({1.496 \; \rm{mol} \; \rm{Ar}}\))
\(\rm{n_{2}}\) = \({\dfrac{({915 \; \cancel{\rm{mL}}}) ({1.496 \; \rm{mol} \; \rm{Ar}})} {({125 \; \cancel{\rm{mL}}})}}\)
\(\rm{n_{2}}\) = \({10.95072 \; \rm{mol} \; \rm{Ar}} ≈ {11.0 \; \rm{mol} \; \rm{Ar}}\)
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, n_{2}, is larger than the initial amount of gas, n_{1}, 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 × 10^{23} 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.\( {11.0 \; \cancel{\rm{mol} \; \rm{Ar}}} \times\) \( \dfrac{6.02 \times 10^{23} \; \rm{atoms} \; \rm{Ar}}{1 \; \cancel{\rm{mol} \; \rm{Ar}}}\) = \( {\text {6.622}} \times {\text{10}^{24}}\) \({\rm{atoms} \; \rm{Ar}}\) ≈ \( {\text {6.62}} \times {\text{10}^{24}}\) \({\rm{atoms} \; \rm{Ar}}\)