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Sagot :
To determine the number of moles of gas remaining in the container after the contraction, we can use the relationship between the volume and the number of moles of gas when pressure and temperature are constant. This is derived from the ideal gas law but simplified for constant pressure and temperature:
[tex]\[ \frac{V_1}{n_1} = \frac{V_2}{n_2} \][/tex]
Here:
- [tex]\(V_1\)[/tex] is the initial volume of the gas, which is 89.6 liters.
- [tex]\(n_1\)[/tex] is the initial number of moles of the gas, which is 4.00 moles.
- [tex]\(V_2\)[/tex] is the reduced volume of the gas, which is 44.8 liters.
- [tex]\(n_2\)[/tex] is the number of moles of gas remaining after contraction, which we need to find.
First, rearrange the equation to solve for [tex]\(n_2\)[/tex]:
[tex]\[ n_2 = n_1 \times \frac{V_2}{V_1} \][/tex]
Substitute the known values into the equation:
[tex]\[ n_2 = 4.00 \text{ moles} \times \frac{44.8 \text{ liters}}{89.6 \text{ liters}} \][/tex]
Simplify the fraction:
[tex]\[ \frac{44.8}{89.6} = 0.5 \][/tex]
Therefore:
[tex]\[ n_2 = 4.00 \text{ moles} \times 0.5 = 2.00 \text{ moles} \][/tex]
So, the number of moles of gas remaining in the reduced container is [tex]\( 2.00 \)[/tex] moles.
[tex]\[ \frac{V_1}{n_1} = \frac{V_2}{n_2} \][/tex]
Here:
- [tex]\(V_1\)[/tex] is the initial volume of the gas, which is 89.6 liters.
- [tex]\(n_1\)[/tex] is the initial number of moles of the gas, which is 4.00 moles.
- [tex]\(V_2\)[/tex] is the reduced volume of the gas, which is 44.8 liters.
- [tex]\(n_2\)[/tex] is the number of moles of gas remaining after contraction, which we need to find.
First, rearrange the equation to solve for [tex]\(n_2\)[/tex]:
[tex]\[ n_2 = n_1 \times \frac{V_2}{V_1} \][/tex]
Substitute the known values into the equation:
[tex]\[ n_2 = 4.00 \text{ moles} \times \frac{44.8 \text{ liters}}{89.6 \text{ liters}} \][/tex]
Simplify the fraction:
[tex]\[ \frac{44.8}{89.6} = 0.5 \][/tex]
Therefore:
[tex]\[ n_2 = 4.00 \text{ moles} \times 0.5 = 2.00 \text{ moles} \][/tex]
So, the number of moles of gas remaining in the reduced container is [tex]\( 2.00 \)[/tex] moles.
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