Problem 85
Question
A balloon is filled with helium gas to a gauge pressure of \(22 \mathrm{mm}\) Hg at \(25^{\circ} \mathrm{C}\). The volume of the gas is \(305 \mathrm{mL},\) and the barometric pressure is \(755 \mathrm{mm}\) Hg. What amount of helium is in the balloon? (Remember that gauge pressure = total pressure - barometric pressure. See page \(452 .\) )
Step-by-Step Solution
Verified Answer
There are approximately 0.0126 moles of helium in the balloon.
1Step 1: Determine Total Pressure
First, we need to find the total pressure inside the balloon using the gauge pressure formula. According to the formula gauge pressure = total pressure - barometric pressure, we rearrange to find total pressure: \[\text{Total pressure} = \text{Gauge pressure} + \text{Barometric pressure}\]Substitute the given values:\[\text{Total pressure} = 22 \, \text{mm Hg} + 755 \, \text{mm Hg} = 777 \, \text{mm Hg}\]
2Step 2: Convert Units
Next, convert the volume from milliliters to liters for ease of calculation with the ideal gas law. Since there are 1000 milliliters in a liter,\[305 \, \text{mL} = 0.305 \, \text{L}\]
3Step 3: Apply the Ideal Gas Law
Use the ideal gas law, \( PV = nRT \), where \(P\) is pressure in atmospheres, \(V\) is volume in liters, \(n\) is moles of gas, \(R\) is the ideal gas constant, and \(T\) is temperature in Kelvin. First, convert the total pressure from mm Hg to atmospheres: \[1 \, \text{atm} = 760 \, \text{mm Hg}\]\[P = \frac{777 \, \text{mm Hg}}{760 \, \text{mm Hg/atm}} \approx 1.022 \text{ atm}\]Convert the temperature from Celsius to Kelvin:\[T = 25^{\circ} \text{C} + 273.15 = 298.15 \text{ K}\]The ideal gas constant \( R = 0.0821 \, \text{L atm mol}^{-1} \text{K}^{-1} \).Substitute into the ideal gas law:\[(1.022 \text{ atm})(0.305 \text{ L}) = n(0.0821 \text{ L atm mol}^{-1} \text{ K}^{-1})(298.15 \text{ K})\]
4Step 4: Solve for Moles
Rearrange the equation from Step 3 to solve for \(n\), the number of moles of helium:\[n = \frac{1.022 \times 0.305}{0.0821 \times 298.15} \approx 0.0126 \text{ moles}\]
Key Concepts
Gauge PressureConversion of UnitsMoles CalculationIdeal Gas Constant
Gauge Pressure
When discussing gases, particularly those in closed systems like a balloon, gauge pressure becomes an essential concept. Gauge pressure is the difference between the total pressure inside a system and the atmospheric pressure outside it.
The formula is given by:
- Gauge pressure = Total pressure - Barometric (or atmospheric) pressure
Conversion of Units
Converting units is a crucial step in any scientific calculation to maintain consistency. In the context of gas laws, converting volumes, pressures, and temperatures into standard units is necessary to apply formulas like the ideal gas law accurately.
In our problem, we converted:
In our problem, we converted:
- Volume: from milliliters (mL) to liters (L), recognizing that 1000 mL = 1 L. This ensures the volume is in the correct unit to use in conjunction with the ideal gas law.
- Pressure: from millimeters of mercury (mm Hg) to atmospheres (atm), using the conversion factor where 1 atm = 760 mm Hg. This is crucial because the ideal gas constant is usually applied in units of atm.
- Temperature: from Celsius to Kelvin, adding 273.15 to the Celsius measurement as 0 Kelvin is the absolute zero temperature and allows us to work with the true scale (Kelvin).
Moles Calculation
Moles are a fundamental unit in chemistry representing a specific number of particles, commonly used to quantify gases. The ideal gas law, written as \( PV = nRT \), is a valuable tool for calculating the number of moles \( n \).
Here's how to isolate \( n \) from the equation:
Here's how to isolate \( n \) from the equation:
- Given: \( P \) (pressure in atm), \( V \) (volume in L), \( T \) (temperature in K)
- Reorganize the formula: \( n = \frac{PV}{RT} \)
Ideal Gas Constant
The ideal gas constant, represented by the symbol \( R \), is a key factor in the ideal gas law equation. It acts as a bridge connecting the pressure, volume, and temperature of a gas to its quantity in moles.
The constant \( R \) is typically given the value:
The constant \( R \) is typically given the value:
- \( R = 0.0821 \, \text{L atm mol}^{-1} \text{K}^{-1} \) when using pressure in atmospheres, volume in liters, and temperature in Kelvin.
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