For ammonia a = 4.17 L^2 atm mol^-2 and b = 0.0371 L mol^-1. Using the low-pressure approximation Z = 1 + (P)/(RT)(b - (a)/(RT)) with R = 0.0821 L atm K^-1 mol^-1, the compressibility factor of ammonia at 350 K and 5.0 atm (three decimals) is

JEE Advanced levelNumerical answerDifficulty 3 of 3rg118

For ammonia $a = 4.17 L^{2} atm mo l^{- 2}$ and $b = 0.0371 L mo l^{- 1}$ . Using the low-pressure approximation $Z = 1 + \frac{P}{RT} (b - \frac{a}{RT})$ with $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ , the compressibility factor of ammonia at $350 K$ and $5.0 atm$ (three decimals) is

Numerical answer type, tolerance 0.002.

RT = 0.0821 \times 350 = 28.735 L atm mo l^{- 1}

a / RT = 4.17/28.735 = 0.14512

. Bracket

= 0.0371 - 0.14512 = - 0.10802

Z = 1 + \frac{5.0}{28.735} \times (- 0.10802) = 1 - 0.01880 = 0.98120

, i.e.

0.981