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Chapter 5: Problem 57

A container is filled with an ideal gas to a pressure of \(40.0\) atm at \(0^{\circ} \mathrm{C}\). a. What will be the pressure in the container if it is heated to \(45^{\circ} \mathrm{C}\) ? b. At what temperature would the pressure be \(1.50 \times 10^{2}\) atm? c. At what temperature would the pressure be \(25.0 \mathrm{~atm} ?\)

Short Answer

Expert verified
a. The pressure in the container at \(45^{\circ} \mathrm{C}\) is approximately \(46.6 \mathrm{~atm}\). b. The temperature at which the pressure would be \(1.50 \times 10^{2} \mathrm{~atm}\) is approximately \(1024.73 \mathrm{~K}\). c. The temperature at which the pressure would be \(25.0 \mathrm{~atm}\) is approximately \(171.22 \mathrm{~K}\).

Step by step solution

01

Convert temperatures to Kelvin

First, we need to convert temperatures given in Celsius to Kelvin. We can do this by adding 273.15 to the Celsius temperature. T1 = 0°C + 273.15 = 273.15 K T2 = 45°C + 273.15 = 318.15 K
02

Calculate the pressure at 45°C (318.15 K)

Use the pressure-temperature relationship formula and the given initial pressure and temperatures to calculate the pressure at the final temperature of 45°C (318.15 K). The formula for the pressure-temperature relationship is: (P1/T1) = (P2/T2) Plugging in the values given in the exercise: (40.0 atm / 273.15 K) = (P2 / 318.15 K) Solve for P2: P2 = (40.0 atm) * (318.15 K / 273.15 K) P2 ≈ 46.6 atm a. The pressure in the container at 45°C is approximately 46.6 atm.
03

Calculate the temperature at 150 atm

Now, we need to calculate the temperature at which the pressure is equal to 150 atm. Using the pressure-temperature relationship formula: (P1/T1) = (P2/T2) Plugging in the values given in the exercise and the new pressure: (40.0 atm / 273.15 K) = (150 atm / T2) Solve for T2: T2 = (150 atm) * (273.15 K / 40.0 atm) T2 ≈ 1024.725 K b. The temperature at which the pressure would be 150 atm is approximately 1024.73 K.
04

Calculate the temperature at 25 atm

Lastly, we will calculate the temperature at which the pressure is equal to 25 atm. Using the pressure-temperature relationship formula: (P1/T1) = (P2/T2) Plugging in the values given in the exercise and the new pressure: (40.0 atm / 273.15 K) = (25 atm / T2) Solve for T2: T2 = (25 atm) * (273.15 K / 40.0 atm) T2 ≈ 171.22 K c. The temperature at which the pressure would be 25 atm is approximately 171.22 K.

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Most popular questions from this chapter

In the "Méthode Champenoise," grape juice is fermented in a wine bottle to produce sparkling wine. The reaction is $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q) \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OH}(a q)+2 \mathrm{CO}_{2}(g) $$ Fermentation of \(750 . \mathrm{mL}\) grape juice (density \(=1.0 \mathrm{~g} / \mathrm{cm}^{3}\) ) is allowed to take place in a bottle with a total volume of \(825 \mathrm{~mL}\) until \(12 \%\) by volume is ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OH}\right)\). Assuming that the \(\mathrm{CO}_{2}\) is insoluble in \(\mathrm{H}_{2} \mathrm{O}\) (actually, a wrong assumption), what would be the pressure of \(\mathrm{CO}_{2}\) inside the wine bottle at \(25^{\circ} \mathrm{C}\) ? (The density of ethanol is \(0.79 \mathrm{~g} / \mathrm{cm}^{3}\).)

A gas sample containing \(1.50 \mathrm{~mol}\) at \(25^{\circ} \mathrm{C}\) exerts a pressure of 400 . torr. Some gas is added to the same container and the temperature is increased to \(50 .{ }^{\circ} \mathrm{C}\). If the pressure increases to 800 . torr, how many moles of gas were added to the container? Assume a constant-volume container.

Calculate the pressure exerted by \(0.5000 \mathrm{~mol} \mathrm{~N}_{2}\) in a 1.0000-L container at \(25.0^{\circ} \mathrm{C}\) a. using the ideal gas law. b. using the van der Waals equation. c. Compare the results.

Ideal gas particles are assumed to be volumeless and to neither attract nor repel each other. Why are these assumptions crucial to the validity of Dalton's law of partial pressures?

A \(15.0\) - \(\mathrm{L}\) tank is filled with \(\mathrm{H}_{2}\) to a pressure of \(2.00 \times 10^{2}\) atm. How many balloons (each \(2.00 \mathrm{~L}\) ) can be inflated to a pressure of \(1.00\) atm from the tank? Assume that there is no temperature change and that the tank cannot be emptied below \(1.00 \mathrm{~atm}\) pressure.
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