For a gas obeying the van der Waals' equation, at the critical point
A
both `((delP)/(delV))_(T)`and `((del^(2)P)/(delV^(2)))_(T)`are zero
B
only `((del^(2)P)/(delV^(2)))_(T)` is not zero
C
`((delP)/(delV))_(T)` is zero but `((del^(2)P)/(delV^(2)))_(T)` is non-zero
D
`((delP)/(delV))_(T)` is non-zero but `((del^(2)P)/(delV^(2)))_(T)` is zero
Text Solution
AI Generated Solution
The correct Answer is:
To solve the question regarding the conditions for a gas obeying the van der Waals equation at the critical point, we can follow these steps:
### Step-by-Step Solution:
1. **Understand the Van der Waals Equation**:
The van der Waals equation for one mole of a gas is given by:
\[
\left(P + \frac{a}{V_m^2}\right)(V_m - b) = RT
\]
where \( P \) is the pressure, \( V_m \) is the molar volume, \( R \) is the universal gas constant, \( T \) is the temperature, and \( a \) and \( b \) are constants specific to the gas.
2. **Identify the Critical Point**:
At the critical point, the gas exhibits unique properties where the distinction between liquid and gas phases disappears. This is characterized by specific conditions of temperature, pressure, and volume.
3. **Derivatives at the Critical Point**:
At the critical point, the first and second derivatives of pressure with respect to volume at constant temperature must be equal to zero:
\[
\left(\frac{\partial P}{\partial V}\right)_T = 0
\]
\[
\left(\frac{\partial^2 P}{\partial V^2}\right)_T = 0
\]
This indicates that the slope of the pressure-volume curve is flat (horizontal) at the critical point.
4. **Implication of the Derivatives**:
The first derivative being zero indicates that there is no change in pressure with a change in volume at the critical point, while the second derivative being zero indicates that the curvature of the pressure-volume graph is also flat at this point.
5. **Conclusion**:
Therefore, for a gas obeying the van der Waals equation at the critical point, both the first and second derivatives of pressure with respect to volume at constant temperature are zero.
### Final Answer:
At the critical point for a gas obeying the van der Waals equation:
\[
\left(\frac{\partial P}{\partial V}\right)_T = 0 \quad \text{and} \quad \left(\frac{\partial^2 P}{\partial V^2}\right)_T = 0
\]
To solve the question regarding the conditions for a gas obeying the van der Waals equation at the critical point, we can follow these steps:
### Step-by-Step Solution:
1. **Understand the Van der Waals Equation**:
The van der Waals equation for one mole of a gas is given by:
\[
\left(P + \frac{a}{V_m^2}\right)(V_m - b) = RT
...
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The critical pressure P_(C) and critical temperature T_(C) for a gas obeying van der Waal's equation are 80 atm at 87^(@) C. molar mass of the gas is 130 g/mole. The compressibility factor for the above gas will be smaller than unity under the following condition: