Question

Elemental sulphur is known to exist as S$_8$ (in vapor state) at 1000 K. At this temperature, if 1 g of sulphur occupies 2.62 L, its pressure (in atm) is (Assume S$_8$ vapors follow the ideal gas equation. Given, R = 0.082 L atm mol$^{-1}$ K$^{-1}$)

• A. 1.0
• B. 0.5
• C. 4.0
• D. 2.0

Hint:

When using the ideal gas law, ensure all units are in standard SI: mol, L, atm, and K.

Answer 1

The Correct Option is B

Use the ideal gas equation: $PV = nRT$
Given: $T = 1000$ K, $V = 2.62$ L, $m = 1$ g, $M_{\text{S}_8} = 8 \times 32 = 256$ g/mol
$\Rightarrow n = \dfrac{1}{256}$ mol
$P = \dfrac{nRT}{V} = \dfrac{1}{256} \cdot \dfrac{0.082 \cdot 1000}{2.62} = \dfrac{82}{256 \cdot 2.62} \approx 0.5$ atm