Question

To minimize the rate of evaporation of the tungsten filament, \(1.4 \times 10^{-5} \mathrm{~mol}\) 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 in the lightbulb is approximately 852.88 Pa when the given values are substituted in the ideal gas law formula, with the temperature converted to Kelvin and the volume converted to m³.

Step-by-step solution

Convert the temperature

First, we need to convert the given temperature from Celsius to Kelvin. The conversion formula is: \(T(K) = T(^\circ C) + 273.15\) Plugging in the given temperature: \(T = 23 + 273.15 = 296.15 \, K\) The temperature in Kelvin is 296.15 K.

Plug in the known values

Now, we have all the values we need to plug into the ideal gas law formula: \(PV = nRT\) Where: - P is the pressure (which we want to find) - V = 600 cm³ (the volume of the lightbulb) - n = \(1.4 \times 10^{-5}\) mol (the number of moles of argon) - R = 8.314 J/mol·K (the ideal gas constant) - T = 296.15 K (the temperature in Kelvin)

Convert the volume

Before plugging in the values, we need to convert the volume from cm³ to m³, as the ideal gas constant is given in J/mol·K which requires the volume in m³. 1 m³ is equal to 1,000,000 cm³. Therefore, to convert the volume, we divide the given volume by 1,000,000: \(\text{Volume in m³} = \frac{600\, \text{cm}³}{1,000,000} = 6 \times 10^{-4}\, \text{m}³\)

Solve for pressure

Now we can plug in all the values into the ideal gas law formula and solve for the pressure P: \(P = \frac{nRT}{V}\) \(P = \frac{(1.4 \times 10^{-5} \,\text{mol})(8.314 \,\text{J/mol} \cdot \text{K})(296.15 \,\text{K})}{6 \times 10^{-4} \,\text{m}^3}\)

Calculate pressure

Perform the calculation: \(P = \frac{(1.4 \times 10^{-5} \,\text{mol})(8.314 \,\text{J/mol} \cdot \text{K})(296.15 \,\text{K})}{6 \times 10^{-4} \,\text{m}^3} = 852.88 \, \text{Pa}\) The pressure of argon in the lightbulb is approximately 852.88 Pa.

Key concepts

Pressure Calculation

Understanding pressure calculation is essential when dealing with problems related to gases. Pressure (\( P \)) can be thought of as the force exerted by a gas per unit area. It is a critical variable in the Ideal Gas Law (\( PV = nRT \)), which relates the pressure of a gas to its volume (\( V \)), the amount in moles (\( n \)), the gas constant (\( R \)), and the temperature (\( T \)).

In the context of the ideal gas equation, solving for pressure requires rearranging the formula to make pressure the subject: \

\( P = \frac{nRT}{V} \)

This is what you’ll calculate when you have the other variables defined. In our lightbulb example with argon, this equation allows us to solve for the pressure exerted by the argon gas within the bulb compartment. For accurate results, ensure that the unit for volume is in cubic meters (\( m^3 \)), matching the unit used in the value of the gas constant (\( R \)).

Temperature Conversion

Temperature conversion is a crucial step in solving most thermodynamics problems, including those involving the Ideal Gas Law. To work with this law, we need to make sure that temperature is measured in Kelvin (\( K \)), the absolute temperature scale that is a standard in scientific calculations.

The conversion from Celsius to Kelvin is straightforward: add 273.15 to your Celsius temperature. So, if you have a temperature of \(23^\text{o}C\), the Kelvin temperature would be: \
\[ T(K) = 23^\text{o}C + 273.15 = 296.15 K \]

It’s important to use Kelvin because the Ideal Gas Law is derived using this scale, and it reflects an absolute measure where 0 K (absolute zero) is theoretically the lowest possible temperature, where particles would have minimal motion.

Moles to Volume Relationship

The moles to volume relationship is described in the Ideal Gas Law through the interdependence between the number of moles of a gas (\( n \)) and its volume (\( V \)). One mole of any gas at standard temperature and pressure (STP) occupies 22.4 liters (\( L \)), and this knowledge helps us relate moles to the volume.

However, the conditions are not always STP, so we often must use the Ideal Gas Law to determine the volume occupied by a gas at different temperatures and pressures. This is shown in the problem where we adjust the volume from centimeters cubed to meters cubed to align with the unit used in the gas constant (R), which is in Joules per mole per Kelvin (\( \frac{J}{mol \times K} \) ).

Converting the volume to a consistent unit allows us to substitute directly into the Ideal Gas Law. For our lightbulb example, converting the volume from cubic centimeters to cubic meters is conducted by dividing the volume in cubic centimeters by one million, because one cubic meter is equal to one million cubic centimeters. Hence, accurately establishing this relationship is key to determining the pressure in the lightbulb.