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Sagot :
Certainly! Let's find the dimensional formula for the constants [tex]\( a \)[/tex] and [tex]\( b \)[/tex] in the van der Waals equation, which is given by:
[tex]\[ \left(p + \frac{a}{v^2}\right)(v - b) = RT \][/tex]
Here's a step-by-step solution:
1. Identify the dimensional formulas of the given quantities:
- [tex]\( p \)[/tex] is the pressure, which has the dimensional formula [tex]\([p] = [M L^{-1} T^{-2}]\)[/tex].
- [tex]\( v \)[/tex] is the volume, with the dimensional formula [tex]\([v] = [L^3]\)[/tex].
- [tex]\( R \)[/tex] is the universal gas constant with the dimensional formula [tex]\([R] = [M L^2 T^{-2} K^{-1}]\)[/tex].
- [tex]\( T \)[/tex] is the temperature, which has the dimensional formula [tex]\([T] = [K]\)[/tex].
2. Analyze each term in the van der Waals equation:
- The term [tex]\( p + \frac{a}{v^2} \)[/tex]:
- [tex]\( p \)[/tex] has the dimensional formula [tex]\([M L^{-1} T^{-2}]\)[/tex].
- [tex]\( \frac{a}{v^2} \)[/tex] should match the dimension of [tex]\( p \)[/tex]. So, [tex]\(\frac{a}{v^2}\)[/tex] must also have the dimensional formula [tex]\([M L^{-1} T^{-2}]\)[/tex].
- Since [tex]\( v \)[/tex] has the dimension [tex]\([L^3]\)[/tex], [tex]\( v^2 \)[/tex] has the dimension [tex]\([L^6]\)[/tex].
Therefore:
[tex]\[ \frac{a}{[L^6]} = [M L^{-1} T^{-2}] \][/tex]
Solving for [tex]\( a \)[/tex]:
[tex]\[ [a] = [M L^{-1} T^{-2}] \times [L^6] = [M L^5 T^{-2}] \][/tex]
3. Determine the dimensional formula for [tex]\( b \)[/tex]:
- The term [tex]\( v - b \)[/tex]:
- [tex]\( v \)[/tex] has the dimensional formula [tex]\([L^3]\)[/tex].
- For the equation to be dimensionally consistent, [tex]\( b \)[/tex] must have the same dimensions as [tex]\( v \)[/tex].
Therefore:
[tex]\[ [b] = [L^3] \][/tex]
4. Summarize the results:
- The dimensional formula for [tex]\( a \)[/tex] is [tex]\([M L^5 T^{-2}]\)[/tex].
- The dimensional formula for [tex]\( b \)[/tex] is [tex]\([L^3]\)[/tex].
Hence, the dimensional formulas are:
- For [tex]\( a \)[/tex]: [tex]\([M L^5 T^{-2}]\)[/tex]
- For [tex]\( b \)[/tex]: [tex]\([L^3]\)[/tex]
[tex]\[ \left(p + \frac{a}{v^2}\right)(v - b) = RT \][/tex]
Here's a step-by-step solution:
1. Identify the dimensional formulas of the given quantities:
- [tex]\( p \)[/tex] is the pressure, which has the dimensional formula [tex]\([p] = [M L^{-1} T^{-2}]\)[/tex].
- [tex]\( v \)[/tex] is the volume, with the dimensional formula [tex]\([v] = [L^3]\)[/tex].
- [tex]\( R \)[/tex] is the universal gas constant with the dimensional formula [tex]\([R] = [M L^2 T^{-2} K^{-1}]\)[/tex].
- [tex]\( T \)[/tex] is the temperature, which has the dimensional formula [tex]\([T] = [K]\)[/tex].
2. Analyze each term in the van der Waals equation:
- The term [tex]\( p + \frac{a}{v^2} \)[/tex]:
- [tex]\( p \)[/tex] has the dimensional formula [tex]\([M L^{-1} T^{-2}]\)[/tex].
- [tex]\( \frac{a}{v^2} \)[/tex] should match the dimension of [tex]\( p \)[/tex]. So, [tex]\(\frac{a}{v^2}\)[/tex] must also have the dimensional formula [tex]\([M L^{-1} T^{-2}]\)[/tex].
- Since [tex]\( v \)[/tex] has the dimension [tex]\([L^3]\)[/tex], [tex]\( v^2 \)[/tex] has the dimension [tex]\([L^6]\)[/tex].
Therefore:
[tex]\[ \frac{a}{[L^6]} = [M L^{-1} T^{-2}] \][/tex]
Solving for [tex]\( a \)[/tex]:
[tex]\[ [a] = [M L^{-1} T^{-2}] \times [L^6] = [M L^5 T^{-2}] \][/tex]
3. Determine the dimensional formula for [tex]\( b \)[/tex]:
- The term [tex]\( v - b \)[/tex]:
- [tex]\( v \)[/tex] has the dimensional formula [tex]\([L^3]\)[/tex].
- For the equation to be dimensionally consistent, [tex]\( b \)[/tex] must have the same dimensions as [tex]\( v \)[/tex].
Therefore:
[tex]\[ [b] = [L^3] \][/tex]
4. Summarize the results:
- The dimensional formula for [tex]\( a \)[/tex] is [tex]\([M L^5 T^{-2}]\)[/tex].
- The dimensional formula for [tex]\( b \)[/tex] is [tex]\([L^3]\)[/tex].
Hence, the dimensional formulas are:
- For [tex]\( a \)[/tex]: [tex]\([M L^5 T^{-2}]\)[/tex]
- For [tex]\( b \)[/tex]: [tex]\([L^3]\)[/tex]
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