Question

The van der Waal's equation \(\left(P + \frac{a}{4V^2}\right)\left(V - \frac{b}{2}\right) = \frac{RT}{2}\) is valid for:

• A. 1 mole of an ideal gas.
• B. 2 moles of a real gas.
• C. $\frac{1}{2}$ mole of an ideal gas.
• D. $\frac{1}{2}$ mole of a real gas.

Hint:

To quickly solve such problems, always compare the coefficients of the constant terms on the right-hand side ($nRT$) and the volume correction term ($nb$). Here, comparing $nRT = \frac{RT}{2}$ immediately gives $n = \frac{1}{2}$ without any tedious calculations.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We need to determine the number of moles ($n$) and the type of gas for which the given modified van der Waals equation is valid.


Step 2: Key Formula or Approach:
The standard van der Waals equation for $n$ moles of a real gas is:
\[ \left(P + \frac{n^2 a}{V^2}\right)(V - nb) = nRT \]
We can compare the coefficients of the given equation to this standard form to solve for $n$.


Step 3: Detailed Explanation:
The given equation is:
\[ \left(P + \frac{a}{4V^2}\right)\left(V - \frac{b}{2}\right) = \frac{RT}{2} \]
Let us compare each term with the standard equation:
1. Comparing the pressure correction term:
\[ \frac{n^2 a}{V^2} = \frac{a}{4V^2} \implies n^2 = \frac{1}{4} \implies n = \frac{1}{2} \]
2. Comparing the volume correction term:
\[ nb = \frac{b}{2} \implies n = \frac{1}{2} \]
3. Comparing the term on the right-hand side:
\[ nRT = \frac{RT}{2} \implies n = \frac{1}{2} \]
Since all comparisons consistently yield $n = \frac{1}{2}$, and the van der Waals equation is uniquely formulated for real (non-ideal) gases, this equation is valid for $\frac{1}{2}$ mole of a real gas.


Step 4: Final Answer:
The correct option is (D).