/*! 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 34 A rigid tank contains \(1.50\) m... [FREE SOLUTION] | 91Ó°ÊÓ

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A rigid tank contains \(1.50\) moles of an ideal gas. Determine the number of moles of gas that must be withdrawn from the tank to lower the pressure of the gas from \(25.0\) atm to \(5.00\) atm. Assume the volume of the tank and the temperature of the gas remain constant during this operation.

Short Answer

Expert verified
1.2 moles of gas must be withdrawn to lower the pressure from \(25.0\) atm to \(5.00\) atm.

Step by step solution

01

Understand the information given

We are given that the initial moles of gas (n1) is 1.50 moles and the initial pressure (P1) is 25.0 atm. The final pressure (P2) is 5.00 atm.
02

Apply the ideal gas law to initial and final state

Note that the volume of the tank and the temperature of the gas remain constant. Therefore, one can use the ideal gas law to establish two equations: Initial state: \(P_1 = n_1 * R * T/V\) Final state: \( P_2 = n_2 * R * T/V\) Divide the final state equation by the initial one. This will give us \(\frac{P_2}{P_1} = \frac{n_2}{n_1}\).
03

Calculate the final moles of gas

Rearrange the equation from Step 2 for \(n_2\), the final moles of gas. This gives \(n_2 = n_1 * \frac{P_2}{P_1}\). Substituting the given values into the equation, we find that \(n_2 = 1.5 * \frac{5}{25} = 0.3\) moles.
04

Determine the moles of gas withdrawn

With the initial and final numbers of moles determined, finding the number of moles of gas withdrawn is a simple subtraction: \(n = n_1 - n_2\). So, \(n = 1.5 - 0.3 = 1.2\) moles of gas must be withdrawn.

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Key Concepts

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

Gas Pressure
Gas pressure is an essential concept when studying gases and their behaviors in various conditions. It is defined as the force exerted by gas particles against the walls of its container per unit area. When these particles move and collide with the container sides, they generate pressure. The speed and frequency of these collisions are related to the chosen variables such as temperature and moles of gas.

In our specific case, the pressure is initially at 25.0 atm and needs to be lowered to 5.0 atm to meet the exercise requirement. Here, the change in pressure is not due to a change in temperature or volume but in the amount of gas. Understanding how moles of gas influence pressure enables us to identify relationships in the ideal gas law, focusing on variables that remain constant.

The pressure of a gas can be significantly impacted by adjusting the moles of gas within a system when the container's volume and temperature do not change. By decreasing the number of moles, we effectively reduce the number of collisions within the container, leading to a decrease in gas pressure.
Moles of Gas
The concept of moles in chemistry helps us quantify the amount of substance in a given sample. One mole is defined as exactly 6.022 x 10^23 particles, which is Avogadro's number. In the context of gases, a mole helps to identify the number of particles present within a container. This, in turn, can define how a gas will behave under specific conditions.

In our exercise, we begin with 1.50 moles of an ideal gas in a tank. To solve for the number of moles that must be withdrawn to lower the pressure, we first need to establish the relationship between pressure and moles when both temperature and volume remain constant. This is simplified using the ideal gas equation, which shows that the moles of gas are directly proportional to pressure under constant conditions.

By employing the relationship derived from the ideal gas law: \( \frac{P_2}{P_1} = \frac{n_2}{n_1} \), we calculate the decreased moles corresponding to a pressure change. The calculated result, 0.3 moles, represents how many moles stay in the tank after the process, illustrating the direct influence that adjusting moles has on pressure. When we calculate the moles needed to be withdrawn, 1.2 moles, it boils down the operation to adjusting the substance amount.
Constant Temperature and Volume
When dealing with gases, many scenarios involve varying combinations of temperature, volume, and pressure according to the ideal gas law. However, in this exercise, both the temperature and volume are held constant, simplifying the relationship between the remaining variables: pressure and moles of gas.

Maintaining a constant temperature ensures that the energy level or kinetic energy of the particles remains unchanged throughout the process. This stability means that temperature doesn't influence the transition from 25.0 atm to 5.0 atm. Similarly, a constant volume implies the space inside the container confines the gas amounts, maintaining constant dimensions.

These steady conditions are crucial since they mean that pressure changes can be directly accounted for by changing the amount of gas alone, without needing to account for influences from fluctuating temperature or expanding/contracting volume. This relationship is neatly outlined using the ideal gas law under constant conditions, which provides a clear path to understanding the interaction between moles of gas and pressure directly.

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Most popular questions from this chapter

Onc mole of oxygen gas is at a pressure of \(6.00 \mathrm{~atm}\) and a temperature of \(27.0^{\circ} \mathrm{C}\). (a) If the gas is heated at constant volume until the pressure triples, what is the final temperature? (b) If the gas is heated so that both the pressure and volume are doubled, what is the final temperature?

The ideal gas law can be recast in terms of the density of a gas. (a) Use dimensional analysis to find an expression for the density \(\rho\) of a gas in terms of the number of moles \(n_{1}\) the volume \(V_{1}\) and the molecular weight \(M\) in kilograms per mole. (b) With the expression found in part (a), show that $$ P=\frac{\rho}{M} R T $$ for an ideal gas. (c) Find the density of the carbon dioxide atmosphere at the surface of Venus, where the pressure is \(90.0\) atm and the temperature is \(7.00 \times 10^{2} \mathrm{~K}\). (d) Would an evacuated steel shell of radius \(1.00 \mathrm{~m}\) and mass \(2.00 \times\) \(10^{2} \mathrm{~kg}\) rise or fall in such an atmosphere? Why?

Before beginning a long trip on a hot day, a driver inflates an automobile tire to a gauge pressure of \(1.80\) atm at \(300 \mathrm{~K}\). At the end of the trip, the gauge pressure has increased to \(2.20\) atm. (a) Assuming the volume has remained constant, what is the temperature of the air inside the tire? (b) What percentage of the original mass of air in the tire should be released so the pressure returns to its original value? Assume the temperature remains at the value found in part (a) and the volume of the tire remains constant as air is released.

For each of the following temperatures, find the equivalent temperature on the indicated scale: (a) \(-273.15^{\circ} \mathrm{C}\) on the Fahrenheit scale, (b) \(98.6 \mathrm{~F}\) on the Celsius scale, and \((c) 100 \mathrm{~K}\) on the Fahrenheit scale.

A \(7.00-\mathrm{L}\) vessel contains \(3.50\) moles of ideal gas at a pressure of \(1.60 \times 10^{6} \mathrm{~Pa}\). Find (a) the temperature of the gas and (b) the average kinetic energy of a gas molecule in the vessel. (c) What additional information would you need if you were asked to find the average speed of a gas molecule?

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