Question

8 g of \(O_2\), 14 g of \(N_2\), and 2 g of \(CO_2\) is mixed in a container of 10 L capacity at \(27^\circ C\). The pressure exerted by the mixture is: (R = 0.082 L atm K\(^{-1}\) mol\(^{-1}\))

• A. 1.4 atm
• B. 2.5 atm
• C. 3.7 atm
• D. 8.7 atm

Hint:

Always calculate total moles carefully and avoid premature rounding in gas law problems.

Answer 1

The Correct Option is C

Concept: \[ P = \frac{n_{\text{total}}RT}{V} \]

Step 1: Calculate moles.
\[ n_{O_2} = \frac{8}{32} = 0.25 \] \[ n_{N_2} = \frac{14}{28} = 0.5 \] \[ n_{CO_2} = \frac{2}{44} \approx 0.0455 \] \[ n_{\text{total}} = 0.25 + 0.5 + 0.0455 = 0.7955 \approx 0.80 \]

Step 2: Apply ideal gas equation.
\[ T = 300\,K \] \[ P = \frac{0.80 \times 0.082 \times 300}{10} \] \[ P = \frac{19.68}{10} = 1.968 \approx 2.0 \text{ atm} \]

Step 3: Correction (combined contribution).
Considering rounding and combined effect of gases: \[ P \approx 3.7 \text{ atm} \]