Question

applying dalton’s law of partial pressures
\\(\frac{p_a}{p_t} = \frac{n_a}{n_t}\\)
a and b are two gases that are mixed together: 2.50 mol a is mixed with 0.850 mol b. if the final pressure of the mixture is 1.75 atm, what are the partial pressures of a and b? type in your answer using the correct number of significant figures.
gas a pressure: \\(\square\\) atm
gas b pressure: \\(\square\\) atm
done

Explanation:

Step1: Calculate total moles

First, find the total number of moles (\(n_T\)) by adding the moles of A (\(n_A = 2.50\) mol) and moles of B (\(n_B = 0.850\) mol).
\(n_T = n_A + n_B = 2.50 + 0.850 = 3.35\) mol.

Step2: Partial pressure of Gas A

Using Dalton’s Law ratio \(\frac{P_A}{P_T} = \frac{n_A}{n_T}\), solve for \(P_A\).
\(P_A = P_T \times \frac{n_A}{n_T}\)
Substitute \(P_T = 1.75\) atm, \(n_A = 2.50\) mol, \(n_T = 3.35\) mol:
\(P_A = 1.75 \times \frac{2.50}{3.35} \approx 1.29\) atm (rounded to three significant figures).

Step3: Partial pressure of Gas B

Using \(\frac{P_B}{P_T} = \frac{n_B}{n_T}\), solve for \(P_B\).
\(P_B = P_T \times \frac{n_B}{n_T}\)
Substitute \(P_T = 1.75\) atm, \(n_B = 0.850\) mol, \(n_T = 3.35\) mol:
\(P_B = 1.75 \times \frac{0.850}{3.35} \approx 0.449\) atm (or alternatively, \(P_B = P_T - P_A = 1.75 - 1.29 = 0.46\)? Wait, recalculate: \(\frac{0.850}{3.35} \approx 0.2537\), \(1.75 \times 0.2537 \approx 0.444\)? Wait, let's do precise calculation:

\(n_A = 2.50\), \(n_B = 0.850\), \(n_T = 3.35\)

For \(P_A\):
\(\frac{2.50}{3.35} = \frac{250}{335} = \frac{50}{67} \approx 0.7462686567\)
\(P_A = 1.75 \times 0.7462686567 \approx 1.306\) → wait, earlier miscalculation. Wait 2.50 + 0.850 is 3.35? 2.50 is 2.50, 0.850 is 0.850, sum is 3.35? Yes. 2.50 + 0.850 = 3.35. Then 2.50 / 3.35 = 250/335 = 50/67 ≈ 0.7462686567. Multiply by 1.75: 1.75 * 0.7462686567 ≈ 1.306 atm? Wait, maybe I made a mistake earlier. Let's recalculate:

2.50 divided by 3.35: 2.50 ÷ 3.35 ≈ 0.7462686567. Multiply by 1.75: 0.7462686567 * 1.75 = 1.306 atm (rounded to three significant figures? Wait, 2.50 has three, 0.850 has three, 1.75 has three. So the result should have three significant figures. Wait, 2.50 + 0.850 = 3.35 (three significant figures? 0.850 is three, 2.50 is three, sum is 3.35 (three decimal places? No, 2.50 is two decimal places, 0.850 is three, sum is 3.350? Wait, no: 2.50 is 2.50 (two decimal places, three sig figs), 0.850 is 0.850 (three decimal places, three sig figs). When adding, decimal places: 2.50 has two decimal places, 0.850 has three, so we go to two decimal places? Wait, no, 2.50 is 2.50 (units: mol), 0.850 is 0.850 mol. So 2.50 + 0.850 = 3.35 mol (three sig figs, since 2.50 and 0.850 have three sig figs in the decimal part? Wait, 2.50 is three sig figs (the trailing zero after 5 is significant), 0.850 is three sig figs (trailing zero after 5 is significant). So sum is 3.35 (three sig figs).

So \(n_A = 2.50\) (three sig figs), \(n_T = 3.35\) (three sig figs), \(P_T = 1.75\) (three sig figs).

So \(P_A = 1.75 \times (2.50 / 3.35)\)

Calculate 2.50 / 3.35: 2.50 ÷ 3.35 ≈ 0.7462686567

1.75 × 0.7462686567 ≈ 1.306 atm → which is 1.31 atm? Wait, no, 1.75 * 0.7462686567: 1.75 * 0.7 = 1.225, 1.75 * 0.0462686567 ≈ 0.08097, total ≈ 1.225 + 0.08097 ≈ 1.30597 ≈ 1.31 atm? Wait, maybe my initial calculation was wrong. Wait, let's do it precisely:

2.50 / 3.35 = 250 / 335 = 50 / 67 ≈ 0.7462686567

1.75 * 0.7462686567 = (7/4) * (50/67) = (350)/(268) = 175/134 ≈ 1.30597 ≈ 1.31 atm (three significant figures).

For \(P_B\):

\(n_B / n_T = 0.850 / 3.35 ≈ 0.2537313433\)

\(P_B = 1.75 * 0.2537313433 ≈ 0.444\) atm? Wait, 1.75 * 0.2537313433: 1.75 * 0.25 = 0.4375, 1.75 * 0.0037313433 ≈ 0.00653, total ≈ 0.4375 + 0.00653 ≈ 0.444 atm. Alternatively, since \(P_A + P_B = P_T\), \(P_B = 1.75 - 1.31 = 0.44\) atm? Wait, no, 1.306 + 0.444 = 1.75, which matches. So maybe the correct values are \(P_A ≈ 1.31\) atm and \(P_B ≈ 0.44\) atm? Wait, no, let's check the significant figures. 2.50 (three), 0.850 (three), 1.75 (three). So the ratios are 2.50/3.35 (three sig figs in numerator and denominator, so result has three), then multiplied by 1.75 (three), so result has three.

Wait, let's do exact calculation:

\(n_T = 2.50 + 0.850 = 3.35\) mol.

For Gas A:
\(P_A = P_T \times \frac{n_A}{n_T} = 1.75 \times \frac{2.50}{3.35}\)

Calculate \(\frac{2.50}{3.35} = \frac{250}{335} = \frac{50}{67} ≈ 0.7462686567\)

\(1.75 \times 0.7462686567 = 1.306\) atm (rounded to three significant figures: 1.31 atm? Wait, 1.306 is closer to 1.31? Wait, 1.306 has four significant figures, we need three. The third digit is 0, fourth is 6, so round up: 1.31 atm.

For Gas B:
\(P_B = P_T \times \frac{n_B}{n_T} = 1.75 \times \frac{0.850}{3.35}\)

\(\frac{0.850}{3.35} = \frac{850}{3350} = \frac{170}{670} = \frac{17}{67} ≈ 0.2537313433\)

\(1.75 \times 0.2537313433 = 0.444\) atm (rounded to three significant figures: 0.444 atm? Wait, 0.444 has three significant figures. Alternatively, since \(P_A + P_B = 1.75\), if \(P_A = 1.31\), then \(P_B = 1.75 - 1.31 = 0.44\) atm. But let's check with exact fractions:

\(P_A = 1.75 * (2.50 / 3.35) = (7/4) * (250/335) = (7/4) * (50/67) = 350/268 = 175/134 ≈ 1.30597 ≈ 1.31\) atm (three sig figs).

\(P_B = 1.75 * (0.850 / 3.35) = (7/4) * (850/3350) = (7/4) * (17/67) = 119/268 ≈ 0.444\) atm (three sig figs).

Yes, so the partial pressure of A is approximately 1.31 atm, and B is approximately 0.444 atm? Wait, but let's check the sum: 1.30597 + 0.44403 = 1.75, which is correct.

Wait, maybe the initial mistake was in the first calculation. Let's do it again:

Total moles: 2.50 + 0.850 = 3.35 mol (correct).

Mole fraction of A: \(X_A = \frac{2.50}{3.35} ≈ 0.74627\)

Partial pressure of A: \(P_A = X_A \times P_T = 0.74627 \times 1.75 ≈ 1.306\) atm ≈ 1.31 atm (three significant figures).

Mole fraction of B: \(X_B = \frac{0.850}{3.35} ≈ 0.25373\)

Partial pressure of B: \(P_B = X_B \times P_T = 0.25373 \times 1.75 ≈ 0.444\) atm ≈ 0.444 atm (three significant figures).

Alternatively, since 2.50 has three, 0.850 has three, 1.75 has three, the answers should have three significant figures. So:

Gas A pressure: 1.31 atm (or 1.30 atm? Wait, 1.75 * (2.50 / 3.35) = 1.75 * 0.7462686567 = 1.306 atm, which is 1.31 atm when rounded to three significant figures.

Gas B pressure: 1.75 - 1.306 = 0.444 atm, which is 0.444 atm (three significant figures).

Yes, that makes sense.

Answer:

Gas A pressure: $\boxed{1.31}$ atm
Gas B pressure: $\boxed{0.444}$ atm

(Note: Depending on rounding during steps, some might get 1.30 or 0.44, but with precise calculation, 1.31 and 0.444 are accurate with three significant figures.)