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 in the lightbulb at ${23}^{\circ }\mathrm{C}$ is approximately $5.66\times {10}^{-4}$ atm.

Step-by-step solution

Convert the temperature to Kelvin

To convert the temperature from Celsius to Kelvin, add 273.15 to the Celsius temperature. T(K) = T(°C) + 273.15 T(K) = 23 + 273.15 T(K) = 296.15 K

Convert volume to liters

To convert the volume from cm³ to liters (L), divide the volume in cm³ by 1000: V(L) = V(cm³) / 1000 V(L) = 600 / 1000 V(L) = 0.6 L

Use Ideal Gas Law to solve for Pressure

Now we have all the values we need to plug into the Ideal Gas Law equation: PV = nRT P = nRT / V To solve for pressure (P), we will use the number of moles (n = 1.4 x 10⁻⁵ mol), volume in liters (V = 0.6 L), temperature in Kelvin (T = 296.15 K), and gas constant (R = 0.0821 L atm/mol K). P = (1.4 x 10⁻⁵ mol) * (0.0821 L atm/mol K) * (296.15 K) / (0.6 L) Now, we can calculate the pressure of argon in the lightbulb: P = 5.66 x 10⁻⁴ atm The pressure of argon in the lightbulb at 23°C is approximately 5.66 x 10⁻⁴ atm.