/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 69 A quantity of \(\mathrm{N}_{2}\)... [FREE SOLUTION] | 91Ó°ÊÓ

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A quantity of \(\mathrm{N}_{2}\) gas originally held at \(4.75\) atm pressure in a 1.00-L container at \(26^{\circ} \mathrm{C}\) is transferred to a 10.0-L container at \(20^{\circ} \mathrm{C}\). A quantity of \(\mathrm{O}_{2}\) gas originally at \(5.25\) atm and \(26^{\circ} \mathrm{C}\) in a 5.00-L container is transferred to this same container. What is the total pressure in the new container?

Short Answer

Expert verified
The total pressure in the new container after transferring both Nâ‚‚ and Oâ‚‚ gases is 2.676 atm.

Step by step solution

01

Convert temperatures to Kelvin

Since we are dealing with temperatures measured in Celsius in the problem, we first need to convert them to Kelvin because the Ideal Gas Law uses Kelvin for temperature. The conversion to Kelvin can be done using the formula K = °C + 273.15. For Nitrogen (N₂): \(T_1\) = 26°C + 273.15 = 299.15 K For Oxygen (O₂): Same temperature, so \(T_1\) = 299.15 K The new temperature for the gases in the new container is 20°C: \(T_2\) = 20°C + 273.15 = 293.15 K
02

Calculate pressure for Nâ‚‚ in the new container

Now we will apply the Ideal Gas Law for the Nâ‚‚ gas: \(P_1V_1/T_1 = P_2V_2/T_2\) Substitute all given values and values from Step 1 for Nâ‚‚: \(4.75\, \text{atm} \cdot 1.00\, \text{L} / 299.15\, \text{K} = P_2 \cdot 10.0\, \text{L} / 293.15\, \text{K}\) Solve for \(P_2\): \(P_2 = 4.75\, \text{atm} \cdot 1.00\, \text{L} \cdot 293.15\, \text{K} / (299.15\, \text{K} \cdot 10.0\, \text{L}) = 0.5155\, \text{atm}\)
03

Calculate pressure for Oâ‚‚ in the new container

Now we will apply the Ideal Gas Law for the Oâ‚‚ gas: \(P_1V_1/T_1 = P_2V_2/T_2\) Substitute all given values and values from Step 1 for Oâ‚‚: \(5.25\, \text{atm} \cdot 5.00\, \text{L} / 299.15\, \text{K} = P_2 \cdot 10.0\, \text{L} / 293.15\, \text{K}\) Solve for \(P_2\): \(P_2 = 5.25\, \text{atm} \cdot 5.00\, \text{L} \cdot 293.15\, \text{K} / (299.15\, \text{K} \cdot 10.0\, \text{L}) = 2.1605\, \text{atm}\)
04

Calculate the total pressure in the new container

Since both Nâ‚‚ and Oâ‚‚ gases occupy the same container, their pressures are additive. To find the total pressure, we will add the pressures obtained in step 2 and step 3: Total pressure = Pressure of Nâ‚‚ + Pressure of Oâ‚‚ Total pressure = 0.5155 atm + 2.1605 atm = 2.676 atm The total pressure in the new container is 2.676 atm.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Temperature conversion to Kelvin
When working with gas laws, it's important to use the Kelvin scale for temperature. This is because gas laws, like the Ideal Gas Law, rely on absolute temperature. The Kelvin scale starts at absolute zero, which makes mathematical calculations more straightforward.

Here's how to convert from Celsius to Kelvin:
  • Identify the temperature in Celsius.
  • Add 273.15 to the Celsius temperature.
By following these steps, you ensure that the calculations remain consistent and error-free. For the given exercise, you needed to convert 26°C (both for \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\)) and 20°C to the Kelvin scale:

- The initial temperatures for \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) were 26°C, which equals 299.15 K.- The final temperature in the container is 20°C, which is equal to 293.15 K.

This simple conversion is a foundational step in applying any gas law so make sure not to skip this step!
Pressure calculations
Pressure plays a critical role in gas law calculations. Understanding how to calculate changes in pressure helps us predict how gases behave under different conditions.

In the exercise, we use the Ideal Gas Law to find the new pressure of each gas in the larger container. The equation used is derived from the Ideal Gas Law, which states that for a given amount of gas, the pressure times the volume divided by the temperature should remain constant if the amount of gas doesn't change:
  • Original pressure: \(P_1\)
  • Original volume: \(V_1\)
  • Original temperature: \(T_1\)
  • New pressure: \(P_2\)
  • New volume: \(V_2\)
  • New temperature: \(T_2\)
The formula used is \(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\). This formula can be rearranged to solve for \(P_2\), the pressure we need to find:

For \(\mathrm{N}_{2}\):- The initial conditions are given as 4.75 atm pressure, in 1.00 L volume, at 299.15 K.- These become 0.5155 atm in the new conditions of 10 L and 293.15 K.For \(\mathrm{O}_{2}\):- Initial conditions were 5.25 atm pressure, in a 5.00 L volume, at 299.15 K.- The new pressure calculated is 2.1605 atm.

By following these calculations, students can gain a deep understanding of how changes in temperature and volume affect pressure.
Gas law applications
The Ideal Gas Law is a valuable tool for explaining and predicting the behavior of gases. It allows us to understand how different variables interact. In this exercise, we see an application of the Ideal Gas Law when combining gases into a single container.

The formula \(PV = nRT\) is a core part of this law. Here we rearrange it to address changes in pressure and volume when the temperature also changes. This means we can predict how the pressure will change when gases are transferred from a smaller to a larger container while adjusting for temperature changes.

Applying this knowledge:
  • We calculated the pressure changes for both gases, \(\mathrm{N}_2\) and \(\mathrm{O}_2\), after transferring them to a new container.
  • This setup also addresses Dalton's Law, which says that the total pressure in a gas mixture is the sum of the partial pressures of the individual gases.
  • The exercise combines these pressures from \(\mathrm{N}_2\) and \(\mathrm{O}_2\) to find the total pressure.
The calculation of 2.676 atm as the total pressure demonstrates the cumulative impact of separate gases in a shared space, allowing for a clearer prediction of real-world gas behavior.

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Most popular questions from this chapter

Calculate the number of molecules in a deep breath of air whose volume is \(2.25 \mathrm{~L}\) at body temperature, \(37^{\circ} \mathrm{C}\), and a pressure of 735 torr.

A mixture of gases contains \(0.75 \mathrm{~mol} \mathrm{~N}_{2}, 0.30 \mathrm{~mol} \mathrm{O}_{2}\) and \(0.15 \mathrm{~mol} \mathrm{CO}_{2}\). If the total pressure of the mixture is \(1.56 \mathrm{~atm}, \mathrm{what}\) is the partial pressure of each component?

The Goodyear blimps, which frequently fly over sporting events, hold approximately \(175,000 \mathrm{ft}^{3}\) of helium. If the gas is at \(23^{\circ} \mathrm{C}\) and \(1.0 \mathrm{~atm}\), what mass of helium is in the blimp?

On a single plot, qualitatively sketch the distribution of molecular speeds for (a) \(\operatorname{Kr}(g)\) at \(-50^{\circ} \mathrm{C}\), (b) \(\mathrm{Kr}(g)\) at \(0{ }^{\circ} \mathrm{C}\), (c) \(\operatorname{Ar}(g)\) at \(0{ }^{\circ} \mathrm{C}\). [Section 10.7]

Which of the following statements best explains why nitrogen gas at STP is less dense than Xe gas at STP? (a) Because Xe is a noble gas, there is less tendency for the Xe atoms to repel one another, so they pack more densely in the gas state. (b) Xe atoms have a higher mass than \(\mathrm{N}_{2}\) molecules. Because both gases at STP have the same number of molecules per unit volume, the Xe gas must be denser. (c) The Xe atoms are larger than \(\mathrm{N}_{2}\) molecules and thus take up a larger fraction of the space occupied by the gas. (d) Because the Xe atoms are much more massive than the \(\mathrm{N}_{2}\) molecules, they move more slowly and thus exert less upward force on the gas container and make the gas appear denser.

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