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A 7.55 L container holds a mixture of two gases at 47 °C. The partial pressures of gas A and gas B, respectively, are 0.345 atm and 0.634 atm. If 0.100 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?

User Mikl X
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2 Answers

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Final answer:

The total pressure will become 0.979 atm.

Step-by-step explanation:

Dalton's law of partial pressures states that the total pressure is equal to the sum of the partial pressures. In this case, we have two gases (A and B) with respective partial pressures of 0.345 atm and 0.634 atm. If 0.100 mol of a third gas is added with no change in volume or temperature, we can calculate the total pressure by simply adding the partial pressures together.

Ptot = PA + PB + PC

Ptot = 0.345 atm + 0.634 atm + 0.0 atm

Ptot = 0.979 atm

Therefore, the total pressure will become 0.979 atm.

User Dierdre
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3 votes

Final answer:

The total pressure in the container after adding the third gas will be approximately 1.326 atm, calculated by summing the initial pressures of the gases and the pressure exerted by the added gas using Dalton's law of partial pressures and the ideal gas law.

Step-by-step explanation:

According to Dalton's law of partial pressures, the total pressure in a container is the sum of the partial pressures of all the gases present. The initial pressures of gas A and gas B are 0.345 atm and 0.634 atm, respectively, giving a total initial pressure of 0.979 atm (sum of 0.345 atm and 0.634 atm).

Using the ideal gas law (PV=nRT), where P is pressure, V is volume, n is moles of gas, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is temperature in Kelvin, we can calculate the pressure that the third gas will exert in the container. We are given that n = 0.100 mol and we know the temperature T in Kelvin is 320.15 K (47 °C = 273.15 + 47).

Now, to find the pressure the third gas will exert, rearrange the ideal gas law to P = nRT/V. Substituting the values into the equation gives P = (0.100 mol)(0.0821 L·atm/mol·K)(320.15 K) / (7.55 L), resulting in a pressure of approximately 0.347 atm for the third gas. To get the final total pressure, add this pressure to the initial pressure: 0.979 atm + 0.347 atm = 1.326 atm.

So, the total pressure in the container after adding the third gas will be approximately 1.326 atm, assuming the gas behaves ideally.

User Dmirkitanov
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