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Chapter 5: Problem 69

A sealed balloon is filled with 1.00 \(\mathrm{L}\) helium at $23^{\circ} \mathrm{C}$ and 1.00 atm . The balloon rises to a point in the atmosphere where the pressure is 220 . torr and the temperature is $-31^{\circ} \mathrm{C}$ . What is the change in volume of the balloon as it ascends from 1.00 atm to a pressure of 220 . torr?

Short Answer

Expert verified
The change in volume of the balloon as it ascends from 1.00 atm to a pressure of 220 torr is approximately \(2.774 L\).

Step by step solution

01

Convert units to appropriate values

First, we will convert temperatures from Celsius to Kelvins, and pressure from torr to atmospheres. Initial temperature: \(T_1 = (23+273) K = 296 K\) Final temperature: \(T_2 = (-31+273) K = 242 K\) Initial pressure: \(P_1 = 1.00~atm\) Final pressure: \(P_2 = \frac{220}{760}~atm\)
02

Apply the Ideal Gas Law

Using the Ideal Gas Law formula: \(PV = nRT\), where: - P is the pressure of the gas - V is the volume of the gas - n is the number of moles of the gas - R is the ideal gas constant (\(0.0821~L\cdot atm\cdot K^{-1}\cdot mol^{-1}\)) - T is the temperature of the gas in Kelvins Since the balloon is sealed and the amount of gas (n) remains constant, we can write the combined gas law formula as: \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\) Rearranging the formula to solve for the final volume, \(V_2\): \(V_2 = \frac{P_1V_1T_2}{P_2T_1}\)
03

Calculate the final volume, \(V_2\)

Using the values from Step 1, plug into the equation from Step 2 to find the final volume: \(V_2 = \frac{(1.00~atm)(1.00~L)(242~K)}{(\frac{220}{760}~ atm)(296~K)}\) \(V_2 \approx 3.774 L\)
04

Calculate the change in volume

Now, we will calculate the change in volume as the balloon ascends: \(\Delta V = V_2 - V_1\) \(\Delta V = 3.774 L - 1.00 L = 2.774 L\) So the change in volume of the balloon as it ascends from 1.00 atm to a pressure of 220 torr is approximately 2.774 L.

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Most popular questions from this chapter

A 2.747 -g sample of manganese metal is reacted with excess \(\mathrm{HCl}\) gas to produce 3.22 \(\mathrm{L} \mathrm{H}_{2}(g)\) at 373 \(\mathrm{K}\) and 0.951 atm and a manganese chloride compound $\left(\mathrm{Mn} \mathrm{Cl}_{x}\right)$. What is the formula of the manganese chloride compound produced in the reaction?

Small quantities of hydrogen gas can be prepared in the laboratory by the addition of aqueous hydrochloric acid to metallic zinc. $$\mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g)$$ Typically, the hydrogen gas is bubbled through water for collection and becomes saturated with water vapor. Suppose 240. mL of hydrogen gas is collected at \(30 .^{\circ} \mathrm{C}\) and has a total pressure of 1.032 atm by this process. What is the partial pressure of hydrogen gas in the sample? How many grams of zinc must have reacted to produce this quantity of hydrogen? (The vapor pressure of water is 32 torr at \(30 .^{\circ} \mathrm{C}\).)

A 1.00 -L gas sample at \(100 .^{\circ} \mathrm{C}\) and 600 . torr contains 50.0\(\%\) helium and 50.0\(\%\) xenon by mass. What are the partial pressures of the individual gases?

A 2.50-L container is filled with 175 g argon. a. If the pressure is 10.0 atm, what is the temperature? b. If the temperature is 225 K, what is the pressure?

A compound containing only \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{N}\) yields the following data: i. Complete combustion of 35.0 \(\mathrm{mg}\) of the compound produced 33.5 \(\mathrm{mg}\) of \(\mathrm{CO}_{2}\) and 41.1 \(\mathrm{mg}\) of $\mathrm{H}_{2} \mathrm{O} .$ ii. A 65.2 -mg sample of the compound was analyzed for nitrogen by the Dumas method (see Exercise \(137 ),\) giving 35.6 \(\mathrm{mL}\) of dry \(\mathrm{N}_{2}\) at \(740 .\) torr and \(25^{\circ} \mathrm{C}\) . iii. The effusion rate of the compound as a gas was measured and found to be \(24.6 \mathrm{mL} / \mathrm{min}\). The effusion rate of argon gas, under identical conditions, is \(24.6 \mathrm{mL} / \mathrm{min}\). What is the molecular formula of the compound?
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