Question

To minimize the rate of evaporation of the tungsten filament, \(1.4 \times 10^{-5}\) mol of argon is placed in a \(600-\mathrm{cm}^{3}\) lightbulb. What is the pressure of argon in the lightbulb at \(23^{\circ} \mathrm{C} ?\)

Short answer

The pressure of argon gas in the lightbulb at \(23^{\circ}C\) is approximately \(5.637 \times 10^{-4} atm\).

Step-by-step solution

Convert units

We need to make sure that all the variables are in the appropriate units for the Ideal Gas Law formula. The volume of the lightbulb needs to be converted from \(600 cm^3\) to liters and the temperature needs to be converted from Celsius to Kelvin. 1. Convert volume from \(cm^3\) to liters: \(600 cm^3 = 0.6 L\) 2. Convert temperature from Celsius to Kelvin: \(23^{\circ} C + 273.15 = 296.15 K\) Now, we have: - \(V = 0.6 L\) - \(T = 296.15 K\)

Plug values into the Ideal Gas Law and solve for pressure

Now that we have the appropriate units, we can plug the given values into the Ideal Gas Law formula to find the pressure: \(PV = nRT\) We want to solve for P, so we can rearrange the formula as follows: \(P = \frac{nRT}{V}\) Next, plug in the values: \(P = \frac{(1.4 \times 10^{-5} mol)(0.0821\frac{L \times atm}{mol \times K})(296.15 K)}{0.6 L}\)

Calculate the pressure

Now we can calculate the pressure of argon gas inside the lightbulb: \(P = \frac{(1.4 \times 10^{-5})(0.0821)(296.15)}{0.6} = 5.637 \times 10^{-4} atm\) The pressure of argon gas in the lightbulb at \(23^{\circ}C\) is approximately \(5.637 \times 10^{-4} atm\).

Key concepts

Pressure Calculation

Pressure calculation is an essential part of applying the Ideal Gas Law. The Ideal Gas Law relates the pressure, volume, number of moles, and temperature of a gas through the formula: \(PV = nRT\). To calculate the pressure \(P\), we rearrange the equation to \(P = \frac{nRT}{V}\). Here, \(n\) stands for the number of moles of the gas, \(R\) is the gas constant, \(T\) is the temperature in Kelvin, and \(V\) is the volume in liters.

When you know the number of moles, the temperature, and the volume of the gas, you simply plug these values into the rearranged equation to calculate the pressure. In the case of our lightbulb problem, substituting the number of moles \(1.4 \times 10^{-5}\, mol\), the volume \(0.6\, L\), and the temperature \(296.15\, K\) along with the gas constant \(0.0821\, \frac{L \times atm}{mol \times K}\) gives us the pressure \(5.637 \times 10^{-4}\, atm\).

The key here is using the correct units and knowing the rearranged Ideal Gas Law formula for calculating pressure.

Temperature Conversion

Temperature conversion is key when dealing with gas laws, especially the Ideal Gas Law. The standard unit for temperature in these calculations is Kelvin. To convert from Celsius to Kelvin, you add 273.15 to the Celsius value.

Here's a simple step-by-step method:

• Start with the temperature in Celsius, for example, \(23^{\circ}C\).
• Add \(273.15\) to the Celsius temperature.
• The converted temperature in Kelvin will be \(296.15\, K\).

Remember, this conversion is crucial because the Ideal Gas Law, along with many scientific equations, operates effectively when universal units like Kelvin are used. This is due to its zero-point basis being absolute zero, unlike Celsius.

Unit Conversion

In any scientific calculation involving the Ideal Gas Law, converting units correctly is a must. As seen in our example problem, converting the volume from cubic centimeters to liters is necessary because the gas constant \(R\) we used has units of liters.

To convert volume from cubic centimeters \( cm^3 \) to liters \( L \), you use the conversion factor:

• \( 1 \) liter is equivalent to \( 1000 \) cubic centimeters.
• Therefore, convert by dividing the volume \( 600 \ cm^3\) by \( 1000 \).
• This gives \( 0.6 \) liters.

Converting to the correct units ensures the accurate application of the Ideal Gas Law. Another important conversion involves maintaining consistency across the whole formula, ensuring that variables like pressure, volume, moles, and temperature are all in the right units for \( R \). Conversions make it easier to solve problems accurately without unit inconsistencies.