Question

ozone molecules in the stratosphere absorb much of the harmful radiation from the sun. how many ozone molecules are present in 2.00 l of air if the temperature is 269 k temperature and pressure is 1.01×10⁻³ atm?
r = 0.081 (\frac{lcdot atm}{molcdot k}) or 8.314 (\frac{kpacdot l}{molcdot k})
○ 5.51×10¹⁹ molecules
○ 9.15×10⁻⁵ molecules
○ 1.82×10²⁰ molecules
○ 1.09×10⁴ molecules

Explanation:

Step1: Identify the gas law

We use the ideal gas law \( PV = nRT \) to find the number of moles (\( n \)) of ozone. Here, \( P \) is pressure, \( V \) is volume, \( R \) is the gas constant, and \( T \) is temperature. Given \( P = 1.01\times10^{-3}\ \text{atm} \) (assuming the pressure unit is atm, as \( R = 0.0821\ \frac{\text{L·atm}}{\text{mol·K}} \) is used), \( V = 2.00\ \text{L} \), \( R = 0.0821\ \frac{\text{L·atm}}{\text{mol·K}} \), and \( T = 269\ \text{K} \).

Step2: Solve for moles (\( n \))

Rearrange the ideal gas law to \( n=\frac{PV}{RT} \). Substitute the values:
\( n=\frac{(1.01\times10^{-3}\ \text{atm})(2.00\ \text{L})}{(0.0821\ \frac{\text{L·atm}}{\text{mol·K}})(269\ \text{K})} \)
First, calculate the numerator: \( (1.01\times10^{-3})(2.00)=2.02\times10^{-3} \).
Then, calculate the denominator: \( (0.0821)(269)\approx22.0849 \).
So, \( n=\frac{2.02\times10^{-3}}{22.0849}\approx9.15\times10^{-5}\ \text{mol} \).

Step3: Convert moles to molecules

Use Avogadro's number (\( N_A = 6.022\times10^{23}\ \text{molecules/mol} \)). The number of molecules (\( N \)) is \( N = n\times N_A \).
Substitute \( n = 9.15\times10^{-5}\ \text{mol} \):
\( N=(9.15\times10^{-5}\ \text{mol})(6.022\times10^{23}\ \text{molecules/mol})\approx5.51\times10^{19}\ \text{molecules} \).

Answer:

5.51×10¹⁹ molecules (corresponding to the first option: 5.51×10¹⁹ molecules)