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An ideal gas is contained in a cylinder with a volume of \(5.0 \times\) \(10^{2} \mathrm{~mL}\) at a temperature of \(30 .^{\circ} \mathrm{C}\) and a pressure of 710 . torr. The gas is then compressed to a volume of \(25 \mathrm{~mL}\), and the temperature is raised to \(820 .{ }^{\circ} \mathrm{C}\). What is the new pressure of the gas?

Short Answer

Expert verified
The new pressure of the gas after being compressed to a volume of \(25\, mL\) and the temperature raised to \(820^{\circ}C\) is approximately \(76.8\, atm\).

Step by step solution

01

Define given quantities and convert units

1. Initial volume, \(V_1 = 5.0 \times 10^2 mL\) 2. Initial temperature, \(T_1 = 30^{\circ}C\) 3. Initial pressure, \(P_1 = 710 \, torr\) 4. Final volume, \(V_2 = 25 \, mL\) 5. Final temperature, \(T_2 = 820^{\circ}C\) We need to convert the temperature to Kelvin and pressure to atm units. 6. \(T_1 = 30 + 273.15 = 303.15\,K\) 7. \(T_2 = 820 + 273.15 = 1093.15\,K\) 8. \(P_1 = \frac{710\, torr}{760\, \frac{torr}{atm}} = 0.9342\, atm\)
02

Use Ideal Gas Law

Since the number of moles (n) and the gas constant (R) are constant, we can create a relationship between the initial and final states of the gas using the Ideal Gas Law: \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\) Where \(P_2\) is the new pressure we are trying to find.
03

Solve for new pressure P2

Rearrange the equation to solve for \(P_2\): \(P_2 = \frac{P_1 V_1 T_2}{V_2 T_1}\) Plug in the given values and solve for \(P_2\): \(P_2 = \frac{(0.9342\, atm)(5.0 \times 10^2\, mL)(1093.15\, K)}{(25\, mL)(303.15\, K)}\)
04

Calculate new pressure P2

Perform the calculations: \(P_2 = \frac{(0.9342)(5.0 \times 10^2)(1093.15)}{(25)(303.15)}\) \(P_2 ≈ 76.8\, atm\) The new pressure of the gas is approximately \(76.8\, atm\).

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Key Concepts

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

Gas Pressure Calculation
Understanding how to calculate gas pressure is fundamental when dealing with problems related to the Ideal Gas Law. The pressure of a gas is a measure of the force that the gas molecules exert on the walls of their container. When working with the Ideal Gas Law, pressure can be represented using different units, such as atmospheres (atm), torr, or pascals (Pa). In the exercise, converting the initial pressure from torr to atm was necessary to use the Ideal Gas Law correctly, as the law involves consistent units.

When compressing a gas or changing its temperature, like in our scenario, the pressure will change according to the Ideal Gas Law, given by the formula:
\[ PV = nRT \]
Where P stands for pressure, V for volume, n for the number of moles, R for the gas constant, and T for temperature in Kelvin. By keeping the amount of gas and the gas constant fixed, the formula simplifies, helping us find the new pressure after changes in volume and temperature.
Temperature Conversion to Kelvin
Temperature plays a crucial role in calculations involving gases. However, it's important to use the correct scale—Kelvin (K)—when applying the Ideal Gas Law. The Kelvin scale is an absolute temperature scale, meaning it starts at absolute zero (the theoretical point where particles have minimal thermal motion).

To convert Celsius to Kelvin, which was necessary in our exercise, you add 273.15 to the Celsius temperature:
\[ T(K) = T(^\textcirc C) + 273.15 \]
So, to convert the initial temperature from 30°C and the final temperature from 820°C to Kelvin, we simply added 273.15 to each, resulting in 303.15 K and 1093.15 K, respectively. Accurate conversion is key to solving chemical problems and especially essential when temperatures are extreme, such as in the given problem.
Chemical Problem Solving
Chemical problem solving often involves a systematic approach to applying chemical laws and principles. The Ideal Gas Law is frequently used to solve problems about gases, like in the given textbook exercise where a gas undergoes compression and a temperature change.

By breaking the problem into smaller, manageable steps, we can systematically find a solution. The steps usually involve identifying the known quantities, converting those quantities to the proper units, applying the relevant chemical laws or formulas, and finally, performing the calculations to obtain the result.

In our exercise, identifying the initial and final states of the gas and then using the Ideal Gas Law allowed us to create a relationship between these states to find the unknown final pressure, demonstrating a practical method of chemical problem solving.

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Most popular questions from this chapter

A bicycle tire is filled with air to a pressure of 75 . psi at a temperature of \(19^{\circ} \mathrm{C}\). Riding the bike on asphalt on a hot day increases the temperature of the tire to \(58^{\circ} \mathrm{C}\). The volume of the tire increases by \(4.0 \%\). What is the new pressure in the bicycle tire?

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 \(25^{\circ} \mathrm{C}\).

Use the following information to identify element \(\mathrm{A}\) and compound \(\mathrm{B}\), then answer questions a and \(\mathrm{b}\). An empty glass container has a mass of \(658.572 \mathrm{~g} .\) It has a mass of \(659.452 \mathrm{~g}\) after it has been filled with nitrogen gas at a pressure of 790 . torr and a temperature of \(15^{\circ} \mathrm{C}\). When the container is evacuated and refilled with a certain element (A) at a pressure of 745 torr and a temperature of \(26^{\circ} \mathrm{C}\), it has a mass of \(660.59 \mathrm{~g}\) Compound \(\mathrm{B}\), a gaseous organic compound that consists of \(85.6 \%\) carbon and \(14.4 \%\) hydrogen by mass, is placed in a stainless steel vessel \((10.68 \mathrm{~L})\) with excess oxygen gas. The vessel is placed in a constant-temperature bath at \(22^{\circ} \mathrm{C}\). The pressure in the vessel is \(11.98 \mathrm{~atm}\). In the bottom of the vessel is a container that is packed with Ascarite and a desiccant. Ascarite is asbestos impregnated with sodium hydroxide; it quantitatively absorbs carbon dioxide: $$ 2 \mathrm{NaOH}(s)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) $$ The desiccant is anhydrous magnesium perchlorate, which quantitatively absorbs the water produced by the combustion reaction as well as the water produced by the above reaction. Neither the Ascarite nor the desiccant reacts with compound \(\mathrm{B}\) or oxygen. The total mass of the container with the Ascarite and desiccant is \(765.3 \mathrm{~g}\) The combustion reaction of compound \(\mathrm{B}\) is initiated by a spark. The pressure immediately rises, then begins to decrease, and finally reaches a steady value of \(6.02 \mathrm{~atm} .\) The stainless steel vessel is carefully opened, and the mass of the container inside the vessel is found to be \(846.7 \mathrm{~g}\). \(\mathrm{A}\) and \(\mathrm{B}\) react quantitatively in a \(1: 1\) mole ratio to form one mole of the single product, gas \(\mathrm{C}\). a. How many grams of \(\mathrm{C}\) will be produced if \(10.0 \mathrm{~L} \mathrm{~A}\) and \(8.60 \mathrm{~L}\) \(\mathrm{B}\) (each at STP) are reacted by opening a stopcock connecting the two samples? b. What will be the total pressure in the system?

Hydrogen cyanide is prepared commercially by the reaction of methane, \(\mathrm{CH}_{4}(g)\), ammonia, \(\mathrm{NH}_{3}\left(\mathrm{~g}\right.\) ), and oxygen, \(\mathrm{O}_{2}(\mathrm{~g})\), at high temperature. The other product is gaseous water. a. Write a chemical equation for the reaction. b. What volume of \(\mathrm{HCN}(\mathrm{g})\) can be obtained from the reaction of \(20.0 \mathrm{~L} \mathrm{CH}_{4}(g), 20.0 \mathrm{~L} \mathrm{NH}_{3}(g)\), and \(20.0 \mathrm{~L} \mathrm{O}_{2}(g) ?\) The volumes of all gases are measured at the same temperature and pressure.

Which noble gas has the smallest density at STP? Explain.

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