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Please explain the entire equation and result

The image shows a series of equations related to calculating work \( W \) done in a thermodynamic process. Let me break down the entire calculation:

### 1. Work Equation
The work done by or on a system in a thermodynamic process is given by:

\[
W = -\int_{V_i}^{V_f} P(V) \, dV
\]

where:
- \( W \) is the work done.
- \( P(V) \) is the pressure as a function of volume \( V \).
- \( V_i \) and \( V_f \) are the initial and final volumes, respectively.

### 2. Specific Case: Pressure as a Power of Volume
Here, the pressure \( P(V) \) is given by \( P(V) = \frac{1}{V^{7/5}} \). So, the work equation becomes:

\[
W = -\int_{1}^{0.249} \frac{dV}{V^{7/5}}
\]

The limits of integration are from \( V_i = 1 \) to \( V_f = 0.249 \).

### 3. Integration of the Function
The integral \( \int \frac{dV}{V^{7/5}} \) can be rewritten as:

\[
\int V^{-7/5} \, dV = \int V^{-1.4} \, dV
\]

The integral of \( V^{-n} \) is \( \frac{V^{-n+1}}{-n+1} \), so:

\[
\int V^{-1.4} \, dV = \frac{V^{-1.4+1}}{-1.4+1} = \frac{V^{-0.4}}{-0.4} = -2.5 \cdot V^{-2/5}
\]

Thus, the integral simplifies to:

\[
W = \frac{5}{2} \left[\frac{1}{V^{2/5}}\right]_1^{0.249}
\]

This expression matches what we evaluated earlier, leading to the result:

\[
\frac{5}{2} \left[\frac{1}{(0.249)^{2/5}} - \frac{1}{(1)^{2/5}}\right] = 1.86 \text{ atm litre}
\]

### 4. Conversion to Joules
To convert from atm-litre to joules, use the conversion factor:

\[
1 \text{ atm litre} = 101.3 \text{ Joules}
\]

So:

\[
1.86 \text{ atm litre} = 1.86 \times 101.3 \text{ Joules} \times 10^{-3} = 188.4 \text{ Joules}
\]

### Summary
The final result of the work done, \( W \), in this thermodynamic process is \( 188.4 \) Joules.

Would you like to explore any particular step further or do you have additional questions?

Here are five related questions you might find useful:
1. What does the negative sign in the work equation signify in a thermodynamic process?
2. How would the work change if the initial and final volumes were different?
3. What are some practical examples where this kind of thermodynamic calculation is used?
4. How would the result differ if pressure followed a different function of volume?
5. What other methods can be used to convert work from atm-litre to Joules?

**Tip:** In thermodynamics, understanding the relationship between pressure, volume, and temperature is key to solving problems related to work and heat.

Understand how to calculate work done in thermodynamics using integrals. Step-by-step explanation with a specific example of calculating work done with pressure as a power of volume.

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