Find the average values of pressure `(p_(m))` varying from 2 to 10 atm if the pressure p and the volume v are related as follows:
`PV^((3)/(2))= 160`

Answer

To find the average value of pressure \( P_m \) varying from 2 to 10 atm, given the relationship \( PV^{\frac{3}{2}} = 160 \), we can follow these steps:

### Step 1: Express Volume \( V \) in terms of Pressure \( P \)
From the equation \( PV^{\frac{3}{2}} = 160 \), we can express \( V \) as:
\[ V^{\frac{3}{2}} = \frac{160}{P} \]
Taking both sides to the power of \( \frac{2}{3} \):
\[ V = \left(\frac{160}{P}\right)^{\frac{2}{3}} = 160^{\frac{2}{3}} P^{-\frac{2}{3}} \]

### Step 2: Set up the formula for the average value of \( P \)
The average value of a function \( f(x) \) over the interval \([a, b]\) is given by:
\[ \text{Average} = \frac{1}{b-a} \int_a^b f(x) \, dx \]
In our case, we want to find the average value of pressure \( P_m \) from \( P = 2 \) to \( P = 10 \):
\[ P_m = \frac{1}{10 - 2} \int_2^{10} P \, dP \]

### Step 3: Substitute \( V \) in the integral
We can substitute \( V \) into the integral:
\[ P_m = \frac{1}{8} \int_2^{10} 160^{\frac{2}{3}} P^{-\frac{2}{3}} \, dP \]

### Step 4: Calculate the integral
Now we can compute the integral:
\[ P_m = \frac{160^{\frac{2}{3}}}{8} \int_2^{10} P^{-\frac{2}{3}} \, dP \]
The integral of \( P^{-\frac{2}{3}} \) is:
\[ \int P^{-\frac{2}{3}} \, dP = \frac{P^{\frac{1}{3}}}{\frac{1}{3}} = 3 P^{\frac{1}{3}} + C \]
Thus, we have:
\[ P_m = \frac{160^{\frac{2}{3}}}{8} \left[ 3 P^{\frac{1}{3}} \right]_2^{10} \]
Calculating the limits:
\[ = \frac{160^{\frac{2}{3}}}{8} \left( 3(10^{\frac{1}{3}}) - 3(2^{\frac{1}{3}}) \right) \]

### Step 5: Simplify the expression
Now we can simplify:
\[ = \frac{160^{\frac{2}{3}}}{8} \cdot 3 \left( 10^{\frac{1}{3}} - 2^{\frac{1}{3}} \right) \]
Calculating \( 160^{\frac{2}{3}} \):
\[ 160^{\frac{2}{3}} = (2^5 \cdot 5)^{\frac{2}{3}} = 2^{\frac{10}{3}} \cdot 5^{\frac{2}{3}} \]
Thus, we have:
\[ = \frac{2^{\frac{10}{3}} \cdot 5^{\frac{2}{3}}}{8} \cdot 3 \left( 10^{\frac{1}{3}} - 2^{\frac{1}{3}} \right) \]
Since \( 8 = 2^3 \):
\[ = \frac{2^{\frac{10}{3} - 3} \cdot 5^{\frac{2}{3}}}{1} \cdot 3 \left( 10^{\frac{1}{3}} - 2^{\frac{1}{3}} \right) \]
\[ = 2^{\frac{1}{3}} \cdot 5^{\frac{2}{3}} \cdot 3 \left( 10^{\frac{1}{3}} - 2^{\frac{1}{3}} \right) \]

### Step 6: Calculate the numerical value
Now we can evaluate the numerical value of \( P_m \):
Using approximate values:
- \( 10^{\frac{1}{3}} \approx 2.15 \)
- \( 2^{\frac{1}{3}} \approx 1.26 \)
- \( 5^{\frac{2}{3}} \approx 3.17 \)
- \( 2^{\frac{1}{3}} \approx 1.26 \)

Substituting these values:
\[ P_m \approx 2^{\frac{1}{3}} \cdot 3.17 \cdot 3 \cdot (2.15 - 1.26) \]
Calculating:
\[ P_m \approx 2^{\frac{1}{3}} \cdot 3.17 \cdot 3 \cdot 0.89 \]
This gives us an approximate average value of \( P_m \).

### Final Answer
Thus, the average value of pressure \( P_m \) is approximately \( 9.7 \) atm.

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