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 \( PV^{\frac{2}{5}} = 0.40 \) at \( V = 1 \, m^3 \), we will follow these steps:

### Step 1: Express Pressure as a Function of Volume
From the equation \( PV^{\frac{2}{5}} = 0.40 \), we can express pressure \( P \) in terms of volume \( V \):
\[ P = \frac{0.40}{V^{\frac{2}{5}}} \]

### Step 2: Differentiate Pressure with Respect to Volume
To find the slope \( \frac{dP}{dV} \), we need to differentiate \( P \) with respect to \( V \):
\[ \frac{dP}{dV} = \frac{d}{dV} \left( 0.40 V^{-\frac{2}{5}} \right) \]

Using the power rule for differentiation, we have:
\[ \frac{dP}{dV} = 0.40 \cdot \left(-\frac{2}{5}\right) V^{-\frac{2}{5}-1} \]
\[ \frac{dP}{dV} = -\frac{0.80}{5} V^{-\frac{7}{5}} \]
\[ \frac{dP}{dV} = -\frac{4}{25} V^{-\frac{7}{5}} \]

### Step 3: Evaluate the Derivative at \( V = 1 \, m^3 \)
Now we need to evaluate \( \frac{dP}{dV} \) at \( V = 1 \):
\[ \frac{dP}{dV} \bigg|_{V=1} = -\frac{4}{25} \cdot 1^{-\frac{7}{5}} \]
Since \( 1^{-\frac{7}{5}} = 1 \):
\[ \frac{dP}{dV} \bigg|_{V=1} = -\frac{4}{25} \]

### Step 4: Convert to Decimal Form
Calculating the decimal value:
\[ -\frac{4}{25} = -0.16 \]

### Conclusion
Thus, the slope of the P-V curve at \( V = 1 \, m^3 \) is:
\[ \boxed{-0.16} \]