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Chapter 5: Problem 12

If you have any two gases in different containers that are the same size at the same pressure and same temperature, what is true about the moles of each gas? Why is this true?

Short Answer

Expert verified
The moles of each gas in the two containers are equal. This is true because, according to the Ideal Gas Law, the relationship between the pressure, volume, temperature, and number of moles must remain constant for two gases given the same conditions. Since both containers share the same pressure, volume, and temperature, the equation \(PV = nRT\) is satisfied, and thus the moles of each gas are equal.

Step by step solution

01

Recognize the given information

Given two containers that are the same size, both containers have the same volume (V). Additionally, both gases have the same pressure (P) and temperature (T).
02

Use the Ideal Gas Law

The Ideal Gas Law is given by the equation PV = nRT, where P is the pressure, V is the volume, n is the number of moles, T is the temperature, and R is the gas constant. By substituting the values given in the problem, we can find the moles of each gas.
03

Compare the two gases

Let the two containers be labeled A and B. Apply the Ideal Gas Law equation to both containers: Container A: \(P_A V_A = n_A R T_A\) Container B: \(P_B V_B = n_B R T_B\) Since both containers have the same pressure, volume, and temperature, we can write: \(P_A = P_B\) \(V_A = V_B\) \(T_A = T_B\)
04

Equate the Ideal Gas Law equations

We can set the two Ideal Gas Law equations equal to each other, since the pressure, volume, and temperature are the same for both containers: \(P_A V_A = n_A R T_A\) \(P_B V_B = n_B R T_B\) \(PV = n_A R T\) \(PV = n_B R T\)
05

Solve for the relationship between moles of each gas

Dividing the two equations, we find the relationship between the moles of each gas: \(\frac{n_A R T}{n_B R T} = \frac{PV}{PV}\) Canceling out the common terms, we are left with: \(\frac{n_A}{n_B} = \frac{PV}{PV}\) \(\frac{n_A}{n_B} = 1\) Thus, the moles of each gas (n_A and n_B) are equal. This is true because, according to the Ideal Gas Law, the relationship between the pressure, volume, temperature, and number of moles must remain constant for two gases given the same conditions.

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Most popular questions from this chapter

The average lung capacity of a human is 6.0 L. How many moles of air are in your lungs when you are in the following situations? a. At sea level \((T=298 \mathrm{K}, P=1.00 \mathrm{atm})\) b. \(10 . \mathrm{m}\) below water \((T=298 \mathrm{K}, P=1.97 \mathrm{atm})\) c. At the top of Mount Everest \((T=200 . \mathrm{K}, P=0.296 \mathrm{atm})\)

Consider separate \(1.0-\mathrm{L}\) samples of \(\mathrm{He}(g)\) and \(\mathrm{UF}_{6}(g),\) both at 1.00 atm and containing the same number of moles. What ratio of temperatures for the two samples would produce the same root mean square velocity?

Consider the following reaction: $$4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Al}_{2} \mathrm{O}_{3}(s)$$ It takes 2.00 L of pure oxygen gas at STP to react completely with a certain sample of aluminum. What is the mass of aluminum reacted?

A hot-air balloon is filled with air to a volume of \(4.00 \times\) \(10^{3} \mathrm{m}^{3}\) at 745 torr and \(21^{\circ} \mathrm{C}\) . The air in the balloon is then heated to \(62^{\circ} \mathrm{C},\) causing the balloon to expand to a volume of \(4.20 \times 10^{3} \mathrm{m}^{3} .\) What is the ratio of the number of moles of air in the heated balloon to the original number of moles of air in the balloon? (Hint: Openings in the balloon allow air to flow in and out. Thus the pressure in the balloon is always the same as that of the atmosphere.)

Silane, SiH, , is the silicon analogue of methane, \(\mathrm{CH}_{4}\) . It is prepared industrially according to the following equations: $$\begin{array}{c}{\mathrm{Si}(s)+3 \mathrm{HCl}(g) \longrightarrow \mathrm{HSiCl}_{3}(l)+\mathrm{H}_{2}(g)} \\ {4 \mathrm{HSiCl}_{3}(l) \longrightarrow \mathrm{SiH}_{4}(g)+3 \mathrm{SiCl}_{4}(l)}\end{array}$$ a. If \(156 \mathrm{mL}\) \(\mathrm{HSiCl}_{3} (d=1.34 \mathrm{g} / \mathrm{mL})\) is isolated when 15.0 \(\mathrm{L}\) \(\mathrm{HCl}\) at 10.0 \(\mathrm{atm}\) and \(35^{\circ} \mathrm{C}\) is used, what is the percent yield of \(\mathrm{HSiCl}_{3} ?\) b. When \(156 \mathrm{HSiCl}_{3}\) is heated, what volume of \(\mathrm{SiH}_{4}\) at 10.0 \(\mathrm{atm}\) and \(35^{\circ} \mathrm{C}\) will be obtained if the percent yield of the reaction is 93.1\(\% ?\)
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