An ideal gas is subjected to a thermodynamic process `PV^(2//5)=0.40` where P is in Pa and V is in `m^(3)`. What is the slope of the P-V curve with volume plotted against x-axis at V=1 `m^(3)`?

Options

A. -1
B. -3.5
C. -2.5
D. None of these

Answer

To find the slope of the P-V curve for the given thermodynamic process \( PV^{2/5} = 0.40 \), we can follow these steps:

### Step 1: Rearranging the equation
We start with the equation:
\[ PV^{2/5} = 0.40 \]
To express \( P \) in terms of \( V \), we rearrange the equation:
\[ P = \frac{0.40}{V^{2/5}} \]

### Step 2: Differentiating \( P \) with respect to \( V \)
Next, we need to find the slope of the curve, which is given by \( \frac{dP}{dV} \). We will differentiate \( P \) with respect to \( V \):
\[ P = 0.40 \cdot V^{-2/5} \]
Using the power rule of differentiation, where \( \frac{d}{dx}(x^n) = n x^{n-1} \), we differentiate:
\[ \frac{dP}{dV} = 0.40 \cdot \left(-\frac{2}{5}\right) V^{-2/5 - 1} \]
This simplifies to:
\[ \frac{dP}{dV} = -\frac{0.80}{5} V^{-7/5} = -\frac{0.16}{V^{7/5}} \]

### Step 3: Evaluating the derivative at \( V = 1 \, m^3 \)
Now, we need to evaluate this derivative at \( V = 1 \, m^3 \):
\[ \frac{dP}{dV} \bigg|_{V=1} = -\frac{0.16}{(1)^{7/5}} = -0.16 \]

### Step 4: Interpreting the result
The slope of the P-V curve at \( V = 1 \, m^3 \) is:
\[ \frac{dP}{dV} = -0.16 \, \text{Pa/m}^3 \]

### Conclusion
Thus, the slope of the P-V curve at \( V = 1 \, m^3 \) is \( -0.16 \) Pa/m³.