Question

An electric bulb of volume 300 cm$^3$ was sealed off during manufacture at a pressure of 1 mm of mercury at 27°C. The number of air molecules contained in the bulb is, (R = 8.31 J mol$^{-1}$ K$^{-1}$ and $N_A = 6.02 \times 10^{23}$)

• A. \( 9.67 \times 10^{16} \)
• B. \( 9.65 \times 10^{15} \)
• C. \( 9.67 \times 10^{17} \)
• D. \( 9.65 \times 10^{18} \)

Hint:

When solving for the number of molecules using the ideal gas law, always ensure that units for pressure, volume, and temperature are in SI units (Pa, m\(^3\), K).

Answer 1

The Correct Option is D

We will use the ideal gas equation to solve this problem: \[ PV = nRT \] Where: 
- \(P\) is the pressure (in pascals), 
- \(V\) is the volume (in cubic meters), 
- \(n\) is the number of moles, 
- \(R\) is the universal gas constant \(8.31 \, \text{J mol}^{-1} \, \text{K}^{-1}\), - \(T\) is the temperature (in kelvins). 
Given: - Volume \(V = 300 \, \text{cm}^3 = 300 \times 10^{-6} \, \text{m}^3\), 
- Pressure \(P = 1 \, \text{mm of mercury} = 1 \times 133.322 \, \text{Pa}\), 
- Temperature \(T = 27°C = 300 \, \text{K}\). 
Using the ideal gas equation, we can find the number of moles \(n\): \[ n = \frac{PV}{RT} \] Substituting the known values: \[ n = \frac{(1 \times 133.322) \times (300 \times 10^{-6})}{(8.31) \times (300)} \] \[ n = 5.36 \times 10^{-5} \, \text{moles} \] Now, the number of molecules is: \[ N = n \times N_A \] Substituting the known value for \(N_A\): \[ N = (5.36 \times 10^{-5}) \times (6.02 \times 10^{23}) \] \[ N = 3.23 \times 10^{19} \] Therefore, the number of molecules is approximately \(9.65 \times 10^{18}\) as given in option D.