Problem 62
Question
A compressed gas cylinder contains \(1.00 \times 10^{3} \mathrm{g}\) argon gas. The pressure inside the cylinder is \(2050 .\) psi (pounds per square inch) at a temperature of \(18^{\circ} \mathrm{C}\). How much gas remains in the cylinder if the pressure is decreased to \(650 .\) psi at a temperature of \(26^{\circ} \mathrm{C} ?\)
Step-by-Step Solution
Verified Answer
After the pressure is decreased to 650 psi and the temperature is increased to 26°C, 332.91 g of argon gas remains in the cylinder.
1Step 1: Convert Celsius to Kelvin
The first step is to convert the given temperatures from Celsius to Kelvin. The formula to convert Celsius to Kelvin is:
T(K) = T(°C) + 273.15
For the initial temperature of 18°C:
T1 = 18 + 273.15 = 291.15 K
For the final temperature of 26°C:
T2 = 26 + 273.15 = 299.15 K
2Step 2: Calculate the number of moles
Next, we need to determine the number of moles (n) of argon from the given mass. The molar mass of argon is approximately 39.95 g/mol. We can find the number of moles by dividing the mass by the molar mass:
n = \(\frac{1.00 \times 10^{3}\text{g}}{39.95\text{g/mol}}\) = 25.03 mol
3Step 3: Convert psi to atm
It is necessary to convert the given pressure values from psi to atmospheres (atm). The conversion factor is:
1 atm = 14.696 psi
For the initial pressure of 2050 psi:
P1 = \(\frac{2050\text{psi}}{14.696\text{psi/atm}}\) = 139.57 atm
For the final pressure of 650 psi:
P2 = \(\frac{650\text{psi}}{14.696\text{psi/atm}}\) = 44.23 atm
4Step 4: Find the ratio of moles in the two conditions
Now we have all the quantities in the ideal gas law in the appropriate units:
P1V = \(n_1\)RT1
P2V = \(n_2\)RT2
Since the volume of the cylinder is constant, we can set the two quantities equal to each other and solve for the ratio of moles in the two conditions:
\(\frac{P_1}{P_2}\) = \(\frac{n_1T_2}{n_2T_1}\)
Plugging in the values, we get:
\(\frac{139.57 \text{ atm}}{44.23 \text{ atm}}\) = \(\frac{25.03 \text{ mol} \times 299.15 \text{ K}}{n_2 \times 291.15 \text{ K}}\)
5Step 5: Solve for the final number of moles
Now we can solve for the final number of moles \(n_2\):
\(n_2\) = \(\frac{25.03 \text{ mol} \times 299.15\text{ K} \times 44.23\text{ atm}}{139.57\text{ atm} \times 291.15\text{ K}}\)
\(n_2\) = 8.333 mol
6Step 6: Calculate the mass of remaining gas
Finally, we can convert the final number of moles back to mass using the molar mass of argon:
mass = \(n_2 \times 39.95\text{g/mol}\)
mass = 8.333 mol * 39.95 g/mol = 332.91 g
So, 332.91 g of argon gas remains in the cylinder after the pressure is decreased to 650 psi and the temperature is increased to 26°C.
Key Concepts
Ideal Gas LawMolar Mass CalculationPressure-Temperature RelationshipConverting Celsius to KelvinConverting psi to atm
Ideal Gas Law
The Ideal Gas Law is a fundamental equation in chemistry that relates the pressure, volume, number of moles, and temperature of an ideal gas. It is expressed as the formula: \[PV = nRT\]Where:
- \(P\) represents the pressure of the gas,
- \(V\) is the volume,
- \(n\) is the number of moles of the gas,
- \(R\) is the ideal gas constant, and
- \(T\) is the temperature in Kelvin.
Molar Mass Calculation
The molar mass is the weight of one mole of a substance, usually expressed in grams per mole (g/mol). It serves as a bridge between the mass of a substance and the number of moles, allowing chemists to convert between these two units. To find out how many moles of a substance you have, you'd divide the mass of the substance by its molar mass.For example, the molar mass of argon is about 39.95 g/mol. If you have \(1.00 \times 10^{3}\) grams of argon, you calculate the number of moles by:\[n = \frac{1.00 \times 10^{3} \text{g}}{39.95 \text{g/mol}}\]Understanding molar mass is crucial for calculations in chemistry, such as when using the Ideal Gas Law or when preparing chemical solutions.
Pressure-Temperature Relationship
The pressure-temperature relationship is an aspect of the Ideal Gas Law that describes how pressure and temperature are directly proportional when the volume and the number of moles remain constant. This means when the temperature of an enclosed gas increases, the pressure also increases if the gas's volume is kept unchanged, and vice versa. This is captured by the fraction of pressures and temperatures in the Ideal Gas Law manipulations, specifically when you compare two states of the gas:\[\frac{P_1}{P_2} = \frac{n_1T_2}{n_2T_1}\]This relationship is valuable in understanding how gases will behave under different thermal conditions, such as in refrigeration systems, car engines, and even when analyzing the problem at hand involving argon gas in a cylinder.
Converting Celsius to Kelvin
Temperature conversion between Celsius and Kelvin is a basic, yet critical task in many chemical calculations. In particular, all temperature values must be in Kelvin when applying the Ideal Gas Law. The conversion is straightforward: to convert Celsius to Kelvin, simply add 273.15 to the Celsius temperature:\[T(K) = T(°C) + 273.15\]Applying this in the context of our problem, to convert the initial temperature of \(18°C\) to Kelvin:\[T1 = 18 + 273.15 = 291.15 K\]And for the final temperature of \(26°C\):\[T2 = 26 + 273.15 = 299.15 K\]This step is often where many students make errors, but remembering that 0°C corresponds to 273.15 K can help avoid mistakes in calculations.
Converting psi to atm
Pressure units often need to be converted to match the standard unit required for equations in chemistry, like atmospheres (atm) for the Ideal Gas Law. Pounds per square inch (psi) is commonly used in the United States but must be converted to atm for these calculations. The conversion factor is:\[1 \text{atm} = 14.696 \text{psi}\]To convert psi to atm, divide the pressure in psi by the conversion factor. In our problem, for the initial pressure:\[P1 = \frac{2050 \text{psi}}{14.696 \text{psi/atm}} = 139.57 \text{atm}\]And for the final pressure:\[P2 = \frac{650 \text{psi}}{14.696 \text{psi/atm}} = 44.23 \text{atm}\]Correct conversion is essential as using incorrect units can lead to erroneous results in chemical calculations.
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