Question

For a vessel at 1832 K containing 10 moles of steam at 50 atm. volume would be (Given : a = 5.46 atm. L²mol^–2 , b = 0.031 L mol^–1) –

• A. 10 L
• B. 20 L
• C. 30 L
• D. 40 L

Answer 1

Answer: (C)

30 L

Answer 2

$\left( 50 + \frac{5.46 \times 10^{2}}{V^{2}} \right)$ (V – 0.31)

= 10 × 0.082 × 1832

Ž V = 30 L