Question

in this lab, you will start off by calculating the number of moles of air molecules inside a flask containing 20 ml of water by cooling the flask down to near 0°c and assuming the vapour pressure of water is negligible at that temperature.
if the volume of the air inside the flask is 147 ml, the gas pressure sensors reads 101.6 kpa and the temperature sensor reads 0.3°c, how many moles of air are present in the flask?
express your answer in mol.
assume r = 8.314 j mol⁻¹ k⁻¹.
answer: 0.0066

Explanation:

Step1: Convert units to SI units

Volume \( V = 147 \, \text{mL} = 147 \times 10^{-3} \, \text{L} = 0.147 \, \text{L} \), but we can also use \( \text{m}^3 \) or \( \text{dm}^3 \). However, since \( R \) is in \( \text{J mol}^{-1} \text{K}^{-1} \), and \( 1 \, \text{Pa} \cdot \text{m}^3 = 1 \, \text{J} \), we convert volume to \( \text{m}^3 \): \( V = 147 \times 10^{-6} \, \text{m}^3 \), pressure \( P = 101.6 \, \text{kPa} = 101.6 \times 10^3 \, \text{Pa} \), temperature \( T = 0.3 + 273.15 = 273.45 \, \text{K} \).

Step2: Use the ideal gas law \( PV = nRT \) to solve for \( n \)

Rearrange the formula: \( n = \frac{PV}{RT} \)
Substitute the values: \( P = 101.6 \times 10^3 \, \text{Pa} \), \( V = 147 \times 10^{-6} \, \text{m}^3 \), \( R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1} \), \( T = 273.45 \, \text{K} \)
\( n = \frac{(101.6 \times 10^3) \times (147 \times 10^{-6})}{8.314 \times 273.45} \)
First, calculate the numerator: \( (101.6 \times 10^3) \times (147 \times 10^{-6}) = 101.6 \times 0.147 = 14.9352 \)
Then, calculate the denominator: \( 8.314 \times 273.45 \approx 8.314 \times 273.45 \approx 2273.0 \)
Now, \( n = \frac{14.9352}{2273.0} \approx 0.00657 \, \text{mol} \approx 0.0066 \, \text{mol} \) (rounded to four decimal places)

Answer:

\( 0.0066 \, \text{mol} \)