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Sagot :
Certainly! Let's solve this problem step-by-step:
### Given:
- The volume of a gas at [tex]\(0{ }^{\circ} C\)[/tex] (273.15 K) is [tex]\(V_0\)[/tex].
- We need to determine the volume of the gas at [tex]\(-1{ }^{\circ} C\)[/tex] (272.15 K) maintaining constant pressure.
### Step-by-Step Solution:
1. Temperature Conversion:
The initial and final temperatures are:
[tex]\[ T_0 = 0{ }^{\circ} C = 273.15 \, \text{K} \][/tex]
[tex]\[ T_{-1} = -1{ }^{\circ} C = 272.15 \, \text{K} \][/tex]
2. Charles's Law Application:
Charles's Law states that the volume of a gas is directly proportional to its temperature (in Kelvin) when pressure is constant:
[tex]\[ \frac{V_0}{T_0} = \frac{V_{-1}}{T_{-1}} \][/tex]
Where:
[tex]\[ V_0 \quad \text{(Initial Volume at } T_0) \][/tex]
[tex]\[ V_{-1} \quad \text{(Final Volume at } T_{-1}) \][/tex]
3. Solving for [tex]\(V_{-1}\)[/tex]:
We rearrange the equation to solve for [tex]\(V_{-1}\)[/tex]:
[tex]\[ V_{-1} = V_0 \times \frac{T_{-1}}{T_0} \][/tex]
Substituting the temperatures:
[tex]\[ V_{-1} = V_0 \times \frac{272.15}{273.15} \][/tex]
4. Calculating the Volume:
By performing the calculation:
[tex]\[ V_{-1} \approx V_0 \times 0.996339 \][/tex]
5. Selecting the Correct Option:
Given the possible options:
- (a) [tex]\(\left(V_0 + \frac{V_0}{273}\right) \, \text{ml}\)[/tex]
- (b) [tex]\(\left(V_0 + \frac{272 V_0}{273}\right) \, \text{ml}\)[/tex]
- (c) [tex]\(\left(V_0 - \frac{V_0}{273}\right) \, \text{ml}\)[/tex]
- (d) [tex]\(\left(\frac{V_0}{273} - V_0\right) \, \text{ml}\)[/tex]
Let's compare these options with the calculated value.
Comparing with each step, we find that the correct option matching our calculation is:
[tex]\[ (c) \left(V_0 - \frac{V_0}{273}\right) \, \text{ml} \][/tex]
6. Verifying the Option:
Let's verify if the chosen option yields the correct value:
[tex]\[ \frac{V_0}{273} \approx 0.003663 \, V_0 \][/tex]
Thus,
[tex]\[ V_0 - \frac{V_0}{273} \approx V_0 \times (1 - 0.003663) = V_0 \times 0.996337 \][/tex]
This approximates very closely to our initial calculation of [tex]\(0.996339 \, V_0\)[/tex]. Therefore, the correct answer is indeed:
[tex]\[ \boxed{\left(V_0 - \frac{V_0}{273}\right) \, \text{ml}} \][/tex]
Thus, the answer is option (c).
### Given:
- The volume of a gas at [tex]\(0{ }^{\circ} C\)[/tex] (273.15 K) is [tex]\(V_0\)[/tex].
- We need to determine the volume of the gas at [tex]\(-1{ }^{\circ} C\)[/tex] (272.15 K) maintaining constant pressure.
### Step-by-Step Solution:
1. Temperature Conversion:
The initial and final temperatures are:
[tex]\[ T_0 = 0{ }^{\circ} C = 273.15 \, \text{K} \][/tex]
[tex]\[ T_{-1} = -1{ }^{\circ} C = 272.15 \, \text{K} \][/tex]
2. Charles's Law Application:
Charles's Law states that the volume of a gas is directly proportional to its temperature (in Kelvin) when pressure is constant:
[tex]\[ \frac{V_0}{T_0} = \frac{V_{-1}}{T_{-1}} \][/tex]
Where:
[tex]\[ V_0 \quad \text{(Initial Volume at } T_0) \][/tex]
[tex]\[ V_{-1} \quad \text{(Final Volume at } T_{-1}) \][/tex]
3. Solving for [tex]\(V_{-1}\)[/tex]:
We rearrange the equation to solve for [tex]\(V_{-1}\)[/tex]:
[tex]\[ V_{-1} = V_0 \times \frac{T_{-1}}{T_0} \][/tex]
Substituting the temperatures:
[tex]\[ V_{-1} = V_0 \times \frac{272.15}{273.15} \][/tex]
4. Calculating the Volume:
By performing the calculation:
[tex]\[ V_{-1} \approx V_0 \times 0.996339 \][/tex]
5. Selecting the Correct Option:
Given the possible options:
- (a) [tex]\(\left(V_0 + \frac{V_0}{273}\right) \, \text{ml}\)[/tex]
- (b) [tex]\(\left(V_0 + \frac{272 V_0}{273}\right) \, \text{ml}\)[/tex]
- (c) [tex]\(\left(V_0 - \frac{V_0}{273}\right) \, \text{ml}\)[/tex]
- (d) [tex]\(\left(\frac{V_0}{273} - V_0\right) \, \text{ml}\)[/tex]
Let's compare these options with the calculated value.
Comparing with each step, we find that the correct option matching our calculation is:
[tex]\[ (c) \left(V_0 - \frac{V_0}{273}\right) \, \text{ml} \][/tex]
6. Verifying the Option:
Let's verify if the chosen option yields the correct value:
[tex]\[ \frac{V_0}{273} \approx 0.003663 \, V_0 \][/tex]
Thus,
[tex]\[ V_0 - \frac{V_0}{273} \approx V_0 \times (1 - 0.003663) = V_0 \times 0.996337 \][/tex]
This approximates very closely to our initial calculation of [tex]\(0.996339 \, V_0\)[/tex]. Therefore, the correct answer is indeed:
[tex]\[ \boxed{\left(V_0 - \frac{V_0}{273}\right) \, \text{ml}} \][/tex]
Thus, the answer is option (c).
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