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Sagot :
To solve this problem, we need to use the combined gas law, which relates the initial and final states of a gas sample without changing the amount of gas. The combined gas law is:
[tex]\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \][/tex]
where:
- [tex]\( P_1 \)[/tex] and [tex]\( P_2 \)[/tex] are the initial and final pressures, respectively,
- [tex]\( V_1 \)[/tex] and [tex]\( V_2 \)[/tex] are the initial and final volumes, respectively,
- [tex]\( T_1 \)[/tex] and [tex]\( T_2 \)[/tex] are the initial and final temperatures, respectively.
### Step-by-Step Solution
1. Identify the given values:
- Initial volume ([tex]\( V_1 \)[/tex]): 735 mL
- Initial pressure ([tex]\( P_1 \)[/tex]): 1.20 atm
- Initial temperature ([tex]\( T_1 \)[/tex]): [tex]\( 112^{\circ} C \)[/tex]
- Final pressure ([tex]\( P_2 \)[/tex]): 643 mmHg
- Final temperature ([tex]\( T_2 \)[/tex]): 281 K
2. Convert the initial temperature from Celsius to Kelvin:
[tex]\[ T_1 = 112^{\circ} \text{C} = 112 + 273.15 = 385.15 \text{ K} \][/tex]
3. Convert the final pressure from mmHg to atm:
[tex]\[ P_2 = 643 \text{ mmHg} \times \left( \frac{1 \text{ atm}}{760 \text{ mmHg}} \right) = 643 \times \frac{1}{760} \approx 0.84605 \text{ atm} \][/tex]
4. Rearrange the combined gas law to solve for [tex]\( V_2 \)[/tex]:
[tex]\[ V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1} \][/tex]
5. Substitute the known values into the formula:
[tex]\[ V_2 = 735 \text{ mL} \times \frac{1.20 \text{ atm}}{0.84605 \text{ atm}} \times \frac{281 \text{ K}}{385.15 \text{ K}} \][/tex]
6. Calculate [tex]\( V_2 \)[/tex]:
[tex]\[ V_2 = 735 \times \frac{1.20}{0.84605} \times \frac{281}{385.15} \][/tex]
7. First, calculate the pressure ratio:
[tex]\[ \frac{1.20}{0.84605} \approx 1.4171 \][/tex]
8. Next, calculate the temperature ratio:
[tex]\[ \frac{281}{385.15} \approx 0.7296 \][/tex]
9. Now, combine these ratios and multiply by the initial volume:
[tex]\[ V_2 = 735 \times 1.4171 \times 0.7296 \approx 760.58 \text{ mL} \][/tex]
Final Result:
The final volume of the gas, when the pressure is 643 mmHg and the temperature is 281 K, is approximately:
[tex]\[ V = 760.58 \text{ mL} \][/tex]
[tex]\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \][/tex]
where:
- [tex]\( P_1 \)[/tex] and [tex]\( P_2 \)[/tex] are the initial and final pressures, respectively,
- [tex]\( V_1 \)[/tex] and [tex]\( V_2 \)[/tex] are the initial and final volumes, respectively,
- [tex]\( T_1 \)[/tex] and [tex]\( T_2 \)[/tex] are the initial and final temperatures, respectively.
### Step-by-Step Solution
1. Identify the given values:
- Initial volume ([tex]\( V_1 \)[/tex]): 735 mL
- Initial pressure ([tex]\( P_1 \)[/tex]): 1.20 atm
- Initial temperature ([tex]\( T_1 \)[/tex]): [tex]\( 112^{\circ} C \)[/tex]
- Final pressure ([tex]\( P_2 \)[/tex]): 643 mmHg
- Final temperature ([tex]\( T_2 \)[/tex]): 281 K
2. Convert the initial temperature from Celsius to Kelvin:
[tex]\[ T_1 = 112^{\circ} \text{C} = 112 + 273.15 = 385.15 \text{ K} \][/tex]
3. Convert the final pressure from mmHg to atm:
[tex]\[ P_2 = 643 \text{ mmHg} \times \left( \frac{1 \text{ atm}}{760 \text{ mmHg}} \right) = 643 \times \frac{1}{760} \approx 0.84605 \text{ atm} \][/tex]
4. Rearrange the combined gas law to solve for [tex]\( V_2 \)[/tex]:
[tex]\[ V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1} \][/tex]
5. Substitute the known values into the formula:
[tex]\[ V_2 = 735 \text{ mL} \times \frac{1.20 \text{ atm}}{0.84605 \text{ atm}} \times \frac{281 \text{ K}}{385.15 \text{ K}} \][/tex]
6. Calculate [tex]\( V_2 \)[/tex]:
[tex]\[ V_2 = 735 \times \frac{1.20}{0.84605} \times \frac{281}{385.15} \][/tex]
7. First, calculate the pressure ratio:
[tex]\[ \frac{1.20}{0.84605} \approx 1.4171 \][/tex]
8. Next, calculate the temperature ratio:
[tex]\[ \frac{281}{385.15} \approx 0.7296 \][/tex]
9. Now, combine these ratios and multiply by the initial volume:
[tex]\[ V_2 = 735 \times 1.4171 \times 0.7296 \approx 760.58 \text{ mL} \][/tex]
Final Result:
The final volume of the gas, when the pressure is 643 mmHg and the temperature is 281 K, is approximately:
[tex]\[ V = 760.58 \text{ mL} \][/tex]
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