Question

3.0x1023 molecules of H2 fill a 21.0 L container at 273 K, what is the pressure in atm ?

Answer 1

Answer: the pressure of the gas in the container is approximately 0.529 atm.

Step-by-step explanation:

To calculate the pressure of
`\( 3.0 * 10^(23) \)` molecules of
`\( H_2 \)` in a 21.0 L container at 273 K, we can use the Ideal Gas Law, which is given by:

`\[ PV = nRT \]`

Where:

`\( P \)` = pressure in atmospheres (atm)

`\( V \)` = volume in liters (L)

`\( n \)` = number of moles of the gas

`\( R \)` = ideal gas constant
`\((0.0821 \ L \cdot atm)/(mol \cdot K) \)`

`\( T \)` = temperature in Kelvin (K)

First, we need to convert the number of molecules to moles using Avogadro's number
`(\( 6.022 * 10^(23) \) molecules/mol)`:

`\[ n = \frac{3.0 * 10^(23) \ \text{molecules}}{6.022 * 10^(23) \ \text{molecules/mol}} \]`

`\[ n \approx 0.498 \ \text{moles} \]`

Now we can plug the values into the Ideal Gas Law and solve for pressure
`\( P \)`:

`\[ P = (nRT)/(V) \]`

`\[ P = \frac{(0.498 \ \text{moles})(0.0821 \ L \cdot atm \cdot mol^(-1) \cdot K^(-1))(273 \ K)}{21.0 \ L} \]`

`\[ P = ((0.498)(0.0821)(273))/(21.0) \]`

`\[ P \approx (11.104)/(21.0) \]`

`\[ P \approx 0.52876 \ \text{atm} \]`