Question

determine the total pressure, in kpa, in a 2.50 - l vessel containing the following mixture of gases at 26.7°c: 0.023 mol he, 0.082 mol h₂, and 0.412 mol ne. report your answer to the nearest whole number. note: r = 8.314 \\(\frac{kpa\cdot l}{mol\cdot k}\\)

Explanation:

Step1: Calculate total moles

Total moles ($n_{total}$) = moles of He + moles of $H_2$ + moles of Ne.
$n_{total} = 0.023 + 0.082 + 0.412 = 0.517$ mol.

Step2: Convert temperature to Kelvin

Temperature ($T$) = $26.7^\circ C + 273.15 = 299.85$ K.

Step3: Apply Ideal Gas Law ($PV = nRT$)

We need to find pressure ($P$), so rearrange: $P = \frac{nRT}{V}$.
Given $V = 2.50$ L, $R = 8.314 \frac{kPa \cdot L}{mol \cdot K}$, $n = 0.517$ mol, $T = 299.85$ K.

Substitute values:
$P = \frac{0.517 \times 8.314 \times 299.85}{2.50}$

Calculate numerator: $0.517 \times 8.314 \approx 4.30$, $4.30 \times 299.85 \approx 1289.7$.

Divide by $V$: $P = \frac{1289.7}{2.50} \approx 515.88$ kPa. Round to nearest whole number: 516 kPa.

Answer:

516