Sure, let’s go through this step-by-step.
Given:
- A monoatomic gas with an adiabatic index [tex]\(\gamma = \frac{5}{3}\)[/tex].
- The gas is compressed adiabatically to [tex]\(\frac{1}{8}\)[/tex] of its original volume.
We aim to find how the pressure of the gas changes.
We know from thermodynamics that for an adiabatic process, the relationship between pressure and volume is described by:
[tex]\[ P_1 V_1^\gamma = P_2 V_2^\gamma \][/tex]
where:
- [tex]\(P_1\)[/tex] and [tex]\(V_1\)[/tex] are the initial pressure and volume respectively.
- [tex]\(P_2\)[/tex] and [tex]\(V_2\)[/tex] are the final pressure and volume respectively.
- [tex]\(\gamma\)[/tex] is the adiabatic index.
Given that [tex]\( V_2 = \frac{1}{8} V_1 \)[/tex], we can rewrite the equation as:
[tex]\[ P_1 V_1^\gamma = P_2 \left( \frac{V_1}{8} \right)^\gamma \][/tex]
Substituting [tex]\(\gamma = \frac{5}{3}\)[/tex] into the equation gives:
[tex]\[ P_1 V_1^{\frac{5}{3}} = P_2 \left( \frac{V_1}{8} \right)^{\frac{5}{3}} \][/tex]
Simplifying [tex]\((\frac{V_1}{8})^{\frac{5}{3}}\)[/tex]:
[tex]\[ \left( \frac{V_1}{8} \right)^{\frac{5}{3}} = V_1^{\frac{5}{3}} \left( \frac{1}{8} \right)^{\frac{5}{3}} \][/tex]
[tex]\[ \left( \frac{1}{8} \right)^{\frac{5}{3}} \][/tex]
So, substituting back into our equation:
[tex]\[ P_1 V_1^{\frac{5}{3}} = P_2 V_1^{\frac{5}{3}} \left( \frac{1}{8} \right)^{\frac{5}{3}} \][/tex]
[tex]\[ \frac{P_2}{P_1} = \left( \frac{1}{8} \right)^{-\frac{5}{3}} \][/tex]
Taking the reciprocal because the exponent is negative:
[tex]\[ \frac{P_2}{P_1} = 8^{\frac{5}{3}} \][/tex]
Now, we need to calculate [tex]\( 8^{\frac{5}{3}} \)[/tex]. The base 8 can be written as [tex]\(2^3\)[/tex]:
[tex]\[ 8 = 2^3 \][/tex]
So:
[tex]\[ 8^{\frac{5}{3}} = (2^3)^{\frac{5}{3}} \][/tex]
[tex]\[ = 2^{3 \cdot \frac{5}{3}} \][/tex]
[tex]\[ = 2^5 \][/tex]
[tex]\[ = 32 \][/tex]
Therefore, the pressure change ratio, [tex]\( \frac{P_2}{P_1} \)[/tex], is 32.
In conclusion, when the monoatomic gas is suddenly compressed to [tex]\(\frac{1}{8}\)[/tex] of its original volume adiabatically, the pressure increases by a factor of 32. This means the new pressure [tex]\( P_2 \)[/tex] will be 32 times the original pressure [tex]\( P_1 \)[/tex].
