Question

If two moles of an ideal gas at 500 K occupies a volume of 41 litres, the pressure of the gas is (R = 0.082 L atm K\textsuperscript{--1} mol\textsuperscript{--1})

• A. 2 atm
• B. 3 atm
• C. 4 atm
• D. 5 atm
• E. 1 atm

Hint:

Look for numerical relationships in KEAM questions! Often $R \times T$ will be a multiple of the volume or pressure given. Here, $0.082 \times 500 = 41$ exactly, which simplifies the calculation significantly.

Answer 1

The Correct Option is A

Concept: The state of an ideal gas is described by the Ideal Gas Equation, which relates pressure, volume, temperature, and the number of moles.
• Equation: $PV = nRT$.
• Variables: $P$ (Pressure), $V$ (Volume), $n$ (Moles), $R$ (Gas constant), $T$ (Absolute temperature in Kelvin).

Step 1: Identify and verify units.
• $n = 2$ moles
• $T = 500$ K
• $V = 41$ L
• $R = 0.082$ L atm K$^{-1}$ mol$^{-1}$ The units are consistent (L, atm, K, mol).

Step 2: Solve for Pressure ($P$). $P = \frac{nRT}{V}$ $P = \frac{2 \times 0.082 \times 500}{41}$ $P = \frac{2 \times (0.082 \times 500)}{41}$ Since $0.082 \times 500 = 41$: $P = \frac{2 \times 41}{41} = 2 \text{ atm}$.