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Chapter 5: Problem 42

A balloon is filled to a volume of \(7.00 \times 10^{2} \mathrm{~mL}\) at a temperature of \(20.0^{\circ} \mathrm{C}\). The balloon is then cooled at constant pressure to a temperature of \(1.00 \times 10^{2} \mathrm{~K}\). What is the final volume of the balloon?

Short Answer

Expert verified
The final volume of the balloon after cooling is approximately \(238.8 \mathrm{~mL}\).

Step by step solution

01

Write down the Charles' Law formula.

Charles' Law states that at constant pressure, the volume of a gas is directly proportional to its temperature in Kelvin. The formula for Charles' Law is: \(V_1/T_1 = V_2/T_2\) where, - \(V_1\) is the initial volume of the gas - \(T_1\) is the initial temperature in Kelvin - \(V_2\) is the final volume of the gas - \(T_2\) is the final temperature in Kelvin
02

Convert the given temperatures to Kelvin.

The initial temperature is given in Celsius, and we need to convert it to Kelvin using the formula: \(T(K) = T(^\circ C) + 273.15\) Initial temperature in Kelvin = \(20.0 + 273.15 = 293.15\ \mathrm{K}\) The final temperature is already given in Kelvin.
03

Substitute the given values into the Charles' Law formula.

Now, we have all the necessary values, and we can substitute them into the Charles' Law formula: \(\frac{7.00 \times 10^{2} \mathrm{~mL}}{293.15 \mathrm{~K}} = \frac{V_2}{1.00 \times 10^{2} \mathrm{~K}}\)
04

Solve for the final volume \(V_2\).

To find the final volume, we can cross-multiply and solve for \(V_2\): \(V_2 = \frac{7.00 \times 10^{2} \mathrm{~mL} \times 1.00 \times 10^{2} \mathrm{~K}}{293.15 \mathrm{~K}}\)
05

Calculate the final volume.

Now, compute the final volume: \(V_2 = \frac{7.00 \times 10^{2} \mathrm{~mL} \times 1.00 \times 10^{2} \mathrm{~K}}{293.15 \mathrm{~K}} = 238.8 \mathrm{~mL}\) The final volume of the balloon after cooling is approximately \(238.8 \mathrm{~mL}\).

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Most popular questions from this chapter

As \(\mathrm{NH}_{3}(g)\) is decomposed into nitrogen gas and hydrogen gas at constant pressure and temperature, the volume of the product gases collected is twice the volume of \(\mathrm{NH}_{3}\) reacted. Explain. As \(\mathrm{NH}_{3}(g)\) is decomposed into nitrogen gas and hydrogen gas at constant volume and temperature, the total pressure increases by some factor. Why the increase in pressure and by what factor does the total pressure increase when reactants are completely converted into products? How do the partial pressures of the product gases compare to each other and to the initial pressure of \(\mathrm{NH}_{3}\) ?

A \(2.00-\mathrm{L}\) sample of \(\mathrm{O}_{2}(g)\) was collected over water at a total pressure of 785 torr and \(25^{\circ} \mathrm{C}\). When the \(\mathrm{O}_{2}(g)\) was dried (water vapor removed), the gas had a volume of \(1.94 \mathrm{~L}\) at \(25^{\circ} \mathrm{C}\) and 785 torr. Calculate the vapor pressure of water at \(25^{\circ} \mathrm{C}\).

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

You have a helium balloon at \(1.00 \mathrm{~atm}\) and \(25^{\circ} \mathrm{C}\). You want to make a hot-air balloon with the same volume and same lift as the helium balloon. Assume air is \(79.0 \%\) nitrogen and \(21.0 \%\) oxygen by volume. The "lift" of a balloon is given by the difference between the mass of air displaced by the balloon and the mass of gas inside the balloon. a. Will the temperature in the hot-air balloon have to be higher or lower than \(25^{\circ} \mathrm{C}\) ? Explain. b. Calculate the temperature of the air required for the hot-air balloon to provide the same lift as the helium balloon at \(1.00\) atm and \(25^{\circ} \mathrm{C}\). Assume atmospheric conditions are \(1.00 \mathrm{~atm}\) and \(25^{\circ} \mathrm{C}\).

An ideal gas at \(7^{\circ} \mathrm{C}\) is in a spherical flexible container having a radius of \(1.00 \mathrm{~cm}\). The gas is heated at constant pressure to \(88^{\circ} \mathrm{C}\). Determine the radius of the spherical container after the gas is heated. [Volume of a sphere \(\left.=(4 / 3) \pi r^{3} .\right]\)
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