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Sagot :
Let's solve this step-by-step:
1. Understand the process:
- This is an isothermal process, meaning the temperature remains constant.
- For an isothermal process involving an ideal gas, the product of pressure and volume remains constant: [tex]\( P_1 \times V_1 = P_2 \times V_2 \)[/tex].
2. Given information:
- The volume of the gas is decreased by 10%, therefore, the new volume [tex]\( V_2 \)[/tex] is [tex]\( 90\% \)[/tex] of the original volume [tex]\( V_1 \)[/tex].
- Mathematically, [tex]\( V_2 = 0.9 \times V_1 \)[/tex].
3. Relate initial and final conditions:
- According to the isothermal process equation, [tex]\( P_1 \times V_1 = P_2 \times V_2 \)[/tex].
- Substitute [tex]\( V_2 \)[/tex] with [tex]\( 0.9 \times V_1 \)[/tex], so [tex]\( P_1 \times V_1 = P_2 \times (0.9 \times V_1) \)[/tex].
4. Solve for the new pressure [tex]\( P_2 \)[/tex]:
- [tex]\( P_1 \times V_1 = P_2 \times 0.9 \times V_1 \)[/tex]
- Divide both sides by [tex]\( 0.9 \times V_1 \)[/tex]:
[tex]\[ P_2 = \frac{P_1}{0.9} \][/tex]
5. Determine the percentage change in pressure:
- Calculate the percentage change in pressure using the formula:
[tex]\[ \text{Percentage change in pressure} = \left( \frac{P_2 - P_1}{P_1} \right) \times 100\% \][/tex]
- Substituting [tex]\( P_2 = \frac{P_1}{0.9} \)[/tex]:
[tex]\[ \text{Percentage change in pressure} = \left( \frac{\frac{P_1}{0.9} - P_1}{P_1} \right) \times 100\% \][/tex]
6. Simplify the calculation:
- Inside the fraction:
[tex]\[ \text{Percentage change in pressure} = \left( \frac{P_1 \left( \frac{1}{0.9} - 1 \right)}{P_1} \right) \times 100\% \][/tex]
- Simplify it:
[tex]\[ \text{Percentage change in pressure} = \left( \frac{\frac{1}{0.9} - 1}{1} \right) \times 100\% \][/tex]
- Since [tex]\( \frac{1}{0.9} = \frac{10}{9} \approx 1.1111 \)[/tex]:
[tex]\[ \frac{10}{9} - 1 = \frac{10}{9} - \frac{9}{9} = \frac{1}{9} \approx 0.1111 \][/tex]
7. Calculate the final percentage change:
[tex]\[ 0.1111 \times 100\% = 11.11\% \][/tex]
Thus, when the volume of a gas is decreased by 10% during an isothermal process, its pressure will increase by approximately 11.11%.
Answer: D. Increase by 11.11%
1. Understand the process:
- This is an isothermal process, meaning the temperature remains constant.
- For an isothermal process involving an ideal gas, the product of pressure and volume remains constant: [tex]\( P_1 \times V_1 = P_2 \times V_2 \)[/tex].
2. Given information:
- The volume of the gas is decreased by 10%, therefore, the new volume [tex]\( V_2 \)[/tex] is [tex]\( 90\% \)[/tex] of the original volume [tex]\( V_1 \)[/tex].
- Mathematically, [tex]\( V_2 = 0.9 \times V_1 \)[/tex].
3. Relate initial and final conditions:
- According to the isothermal process equation, [tex]\( P_1 \times V_1 = P_2 \times V_2 \)[/tex].
- Substitute [tex]\( V_2 \)[/tex] with [tex]\( 0.9 \times V_1 \)[/tex], so [tex]\( P_1 \times V_1 = P_2 \times (0.9 \times V_1) \)[/tex].
4. Solve for the new pressure [tex]\( P_2 \)[/tex]:
- [tex]\( P_1 \times V_1 = P_2 \times 0.9 \times V_1 \)[/tex]
- Divide both sides by [tex]\( 0.9 \times V_1 \)[/tex]:
[tex]\[ P_2 = \frac{P_1}{0.9} \][/tex]
5. Determine the percentage change in pressure:
- Calculate the percentage change in pressure using the formula:
[tex]\[ \text{Percentage change in pressure} = \left( \frac{P_2 - P_1}{P_1} \right) \times 100\% \][/tex]
- Substituting [tex]\( P_2 = \frac{P_1}{0.9} \)[/tex]:
[tex]\[ \text{Percentage change in pressure} = \left( \frac{\frac{P_1}{0.9} - P_1}{P_1} \right) \times 100\% \][/tex]
6. Simplify the calculation:
- Inside the fraction:
[tex]\[ \text{Percentage change in pressure} = \left( \frac{P_1 \left( \frac{1}{0.9} - 1 \right)}{P_1} \right) \times 100\% \][/tex]
- Simplify it:
[tex]\[ \text{Percentage change in pressure} = \left( \frac{\frac{1}{0.9} - 1}{1} \right) \times 100\% \][/tex]
- Since [tex]\( \frac{1}{0.9} = \frac{10}{9} \approx 1.1111 \)[/tex]:
[tex]\[ \frac{10}{9} - 1 = \frac{10}{9} - \frac{9}{9} = \frac{1}{9} \approx 0.1111 \][/tex]
7. Calculate the final percentage change:
[tex]\[ 0.1111 \times 100\% = 11.11\% \][/tex]
Thus, when the volume of a gas is decreased by 10% during an isothermal process, its pressure will increase by approximately 11.11%.
Answer: D. Increase by 11.11%
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