Question

1. a mixture of neon and argon gases exerts a total pressure of 2.39 atm. the partial pressure of the neon alone is 1.84 atm, what is the partial pressure of the argon gas in kpa? 3. a 5.0-liter container at 20.0°c has 4 gases pumped in. the total pressure of the gases is 4.80 atm. if the pressure of the first gas is 1.20 atm, and the pressure of the second gas is 0.490 atm, the pressure of the third gas is 0.780 atm, what is the pressure of the fourth gas in atmospheres?

Explanation:

Problem 2

Step1: Recall Dalton's Law

Dalton's Law: \( P_{\text{total}} = P_1 + P_2 \), where \( P_{\text{total}} \) is total pressure, \( P_1, P_2 \) are partial pressures.
Given \( P_{\text{total}} = 2.39 \, \text{atm} \), \( P_{\text{neon}} = 1.84 \, \text{atm} \). Find \( P_{\text{argon}} \).

Step2: Solve for \( P_{\text{argon}} \)

\( P_{\text{argon}} = P_{\text{total}} - P_{\text{neon}} = 2.39 - 1.84 = 0.55 \, \text{atm} \).

Step3: Convert atm to kPa

1 atm = 101.325 kPa, so \( P_{\text{argon}} = 0.55 \times 101.325 \approx 55.73 \, \text{kPa} \).

Step1: Recall Dalton's Law

Dalton's Law: \( P_{\text{total}} = P_1 + P_2 + P_3 + P_4 \), where \( P_{\text{total}} = 4.80 \, \text{atm} \), \( P_1 = 1.20 \, \text{atm} \), \( P_2 = 0.490 \, \text{atm} \), \( P_3 = 0.780 \, \text{atm} \). Find \( P_4 \).

Step2: Rearrange formula for \( P_4 \)

\( P_4 = P_{\text{total}} - (P_1 + P_2 + P_3) \).

Step3: Calculate sum of known pressures

\( P_1 + P_2 + P_3 = 1.20 + 0.490 + 0.780 = 2.47 \, \text{atm} \).

Step4: Solve for \( P_4 \)

\( P_4 = 4.80 - 2.47 = 2.33 \, \text{atm} \).

Answer:

The partial pressure of argon is approximately \( 55.7 \, \text{kPa} \) (or 56 kPa, depending on rounding).

Problem 3