Certainly! Let's work through this problem step by step.
We start with the information we're given:
- Initial moles of gas ([tex]\( n_1 \)[/tex]): 3.75 moles
- Initial volume ([tex]\( V_1 \)[/tex]): 20.0 liters
- The container shrinks by 8.0 liters to a new volume ([tex]\( V_2 \)[/tex]).
First, calculate the new volume of the container:
[tex]\[ V_2 = V_1 - 8.0 \text{ L} \][/tex]
[tex]\[ V_2 = 20.0 \text{ L} - 8.0 \text{ L} \][/tex]
[tex]\[ V_2 = 12.0 \text{ L} \][/tex]
Next, we use the fact that the pressure and temperature remain constant. According to Boyle’s Law and the Ideal Gas Law, the ratio of the number of moles to the volume should remain constant if the pressure and temperature do not change. This is expressed as:
[tex]\[ \frac{n_1}{V_1} = \frac{n_2}{V_2} \][/tex]
We can rearrange this equation to solve for the final number of moles [tex]\( n_2 \)[/tex]:
[tex]\[ n_2 = n_1 \times \frac{V_2}{V_1} \][/tex]
Substitute the known values into the equation:
[tex]\[ n_2 = 3.75 \text{ moles} \times \frac{12.0 \text{ L}}{20.0 \text{ L}} \][/tex]
Simplify the fraction:
[tex]\[ \frac{12.0 \text{ L}}{20.0 \text{ L}} = 0.6 \][/tex]
So we have:
[tex]\[ n_2 = 3.75 \text{ moles} \times 0.6 \][/tex]
[tex]\[ n_2 = 2.25 \text{ moles} \][/tex]
Therefore, the number of moles of gas in the contracted container is:
[tex]\[ n_2 = 2.25 \text{ moles} \][/tex]
Thus, the final answer is:
[tex]\[ n_2 = 2.25 \text{ mol} \][/tex]
