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Chapter 6: Problem 25

A constant-volume vessel contains \(12.5 \mathrm{g}\) of a gas at 21 ^ C. If the pressure of the gas is to remain constant as the temperature is raised to \(210^{\circ} \mathrm{C}\), how many grams of gas must be released?

Short Answer

Expert verified
No amount of gas needs to be released. Actually, about 7.88 g of gas would have to be added.

Step by step solution

01

Understanding the problem

The problem has given that the volume and pressure of the gas will remain constant and the temperature is changed. We need to find out how many grams of gas must be released to maintain the constant pressure when the temperature is raised.
02

Convert the temperatures from Celsius to Kelvin

The absolute temperature scale (Kelvin) should be used when dealing with gas laws. So, the initial temperature is \(T_1 = 21^{\circ} \mathrm{C} + 273.15 = 294.15 \mathrm{K}\) and the final temperature is \(T_2 = 210^{\circ} \mathrm{C} + 273.15 = 483.15 \mathrm{K}\)
03

Apply gas law equation

The amount of gas (in moles) is directly proportional to its absolute temperature. Let's call the initial mass of the gas \(m_1\) and the final mass \(m_2\). We can write \(\frac{m_1}{T_1} = \frac{m_2}{T_2}\)
04

Calculate the final mass

We can rearrange for \(m_2\) to find out the final mass necessary to keep the pressure constant: \(m_2 = m_1 * \frac{T_2}{T_1}\). The initial mass is given as 12.5 g, so \(m_2 = 12.5 \mathrm{g} * \frac{483.15 \mathrm{K}}{294.15 \mathrm{K}} = 20.38 \mathrm{g}\)
05

Find out the mass of gas to be released

The mass to be released will be the mass we start with (12.5 g) minus the mass we need to have at the end (20.38 g), which is negative. This means that instead of having to release gas, we should have added gas. So no gas has to be released to maintain the pressure.

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Most popular questions from this chapter

A 132.10 mL glass vessel weighs 56.1035 g when evacuated and \(56.2445 \mathrm{g}\) when filled with the gaseous hydrocarbon acetylene at \(749.3 \mathrm{mmHg}\) and \(20.02^{\circ} \mathrm{C}\) What is the molar mass of acetylene? What conclusion can you draw about its molecular formula?

What is the partial pressure of \(\mathrm{Cl}_{2}(\mathrm{g}),\) in millimeters of mercury, at \(0.00^{\circ} \mathrm{C}\) and 1.00 atm in a gaseous mixture that consists of \(46.5 \% \mathrm{N}_{2}, 12.7 \% \mathrm{Ne},\) and \(40.8 \%\) \(\mathrm{Cl}_{2},\) by mass?

In order for a gas-filled balloon to rise in air, the density of the gas in the balloon must be less than that of air. (a) Consider air to have a molar mass of \(28.96 \mathrm{g} / \mathrm{mol}\) determine the density of air at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{atm},\) in g/L. (b) Show by calculation that a balloon filled with carbon dioxide at \(25^{\circ} \mathrm{C}\) and 1 atm could not be expected to rise in air at \(25^{\circ} \mathrm{C}\)

If the van der Waals equation is solved for volume, a cubic equation is obtained. (a) Derive the equation below by rearranging equation (6.26). \(V^{3}-n\left(\frac{R T+b P}{P}\right) V^{2}+\left(\frac{n^{2} a}{P}\right) V-\frac{n^{3} a b}{P}=0\) (b) What is the volume, in liters, occupied by \(185 \mathrm{g}\) \(\mathrm{CO}_{2}(\mathrm{g})\) at a pressure of \(125 \mathrm{atm}\) and \(286 \mathrm{K} ?\) For \(\mathrm{CO}_{2}(\mathrm{g})\) \(a=3.61 \mathrm{L}^{2} \mathrm{atm} \mathrm{mol}^{-2}\) and \(b=0.0429 \mathrm{Lmol}^{-1}\) [Hint: Use the ideal gas equation to obtain an estimate of the volume. Then refine your estimate, either by trial and error, or using the method of successive approximations. See Appendix A, pages A5-A6, for a description of the method of successive approximations.

Consider the statements (a) to (e) below. Assume that \(\mathrm{H}_{2}(\mathrm{g})\) and \(\mathrm{O}_{2}(\mathrm{g})\) behave ideally. State whether each of the following statements is true or false. For each false statement, explain how you would change it to make it a true statement. (a) Under the same conditions of temperature and pressure, the average kinetic energy of \(\mathrm{O}_{2}\) molecules is less than that of \(\mathrm{H}_{2}\) molecules. (b) Under the same conditions of temperature and pressure, \(\mathrm{H}_{2}\) molecules move faster, on average, than \(\mathrm{O}_{2}\) molecules. (c) The volume of \(1.00 \mathrm{mol}\) of \(\mathrm{H}_{2}(\mathrm{g})\) at \(25.0^{\circ} \mathrm{C}\) 1.00 atm is \(22.4 \mathrm{L}\) (d) The volume of \(2.0 \mathrm{g} \mathrm{H}_{2}(\mathrm{g})\) is equal to the volume of \(32.0 \mathrm{g} \mathrm{O}_{2}(\mathrm{g}),\) at the same temperature and pressure. (e) In a mixture of \(\mathrm{H}_{2}\) and \(\mathrm{O}_{2}\) gases, with partial pressures \(P_{\mathrm{H}_{2}}\) and \(P_{\mathrm{O}_{2}^{\prime}}\) respectively, the total pressure is the larger of \(P_{\mathrm{H}_{2}}\) and \(P_{\mathrm{O}_{2}}\).
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