Which of the following statements is(are) true? a. If the number of moles of a gas is doubled, the volume will double, assuming the pressure and temperature of the gas remain constant. b. If the temperature of a gas increases from \(25^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\) the volume of the gas would double, assuming that the pressure and the number of moles of gas remain constant. c. The device that measures atmospheric pressure is called a barometer. d. If the volume of a gas decreases by one half, then the pressure would double, assuming that the number of moles and the temperature of the gas remain constant.

Short Answer

Expert verified
The true statements are: a, c, and d.

Step by step solution

01

Statement a: Doubling the number of moles, volume will double

Using the Ideal Gas Law, \(PV = nRT\), assume the pressure and temperature are constant. If we double the number of moles, we now have \(P(2V) = 2nRT\). Divide this equation by the original ideal gas equation, \(PV = nRT\), to get \(\frac{2V}{V} = \frac{2nRT}{nRT}\), which simplifies to 2 = 2. This shows that statement a is true.
02

Statement b: Doubling the temperature, volume will double

Again, using the Ideal Gas Law, \(PV = nRT\), assume the pressure and the number of moles are constant. Convert the given temperatures from Celsius to Kelvin by adding 273.15. So, we have that T1 = 273.15 K + 25 K = 298.15 K, and T2 = 273.15 K + 50 K = 323.15 K. Now let's calculate the ratio of the volumes with the given temperatures: \(\frac{V_2}{V_1} = \frac{nRT_2}{nRT_1}\). The number of moles and the gas constant R are constant, so we get \(\frac{V_2}{V_1} = \frac{T_2}{T_1} = \frac{323.15 K}{298.15 K} \approx 1.08\). Since the volume ratio is not equal to 2, statement b is false.
03

Statement c: Barometer measures atmospheric pressure

A barometer is an instrument that is used to measure atmospheric pressure. This statement is just a definition and is true.
04

Statement d: Halving the volume, pressure will double

From the Ideal Gas Law, \(PV = nRT\), let's assume the number of moles and the temperature are constant. If the volume decreases by half, then the equation becomes \(P \times \frac{V}{2} = nRT\). We can rearrange this equation to find the relationship between the original pressure, P, and the new pressure, P': \( P' = \frac{2nRT}{V} = 2P\). This shows that statement d is true as well. In conclusion, the true statements are: a, c, and d.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Calculate the pressure exerted by \(0.5000\) mole of \(\mathrm{N}_{2}\) in a \(10.000-\mathrm{L}\) container at \(25.0^{\circ} \mathrm{C}\). a. using the ideal gas law. b. using the van der Waals equation. c. Compare the results. d. Compare the results with those in Exercise 115.

Consider an equimolar mixture (equal number of moles) of two diatomic gases \(\left(A_{2} \text { and } B_{2}\right)\) in a container fitted with a piston. The gases react to form one product (which is also a gas) with the formula \(A_{x} B_{y}\). The density of the sample after the reaction is complete (and the temperature returns to its original state) is \(1.50\) times greater than the density of the reactant mixture. a. Specify the formula of the product, and explain if more than one answer is possible based on the given data. b. Can you determine the molecular formula of the product with the information given or only the empirical formula?

The oxides of Group \(2 \mathrm{A}\) metals (symbolized by M here) react with carbon dioxide according to the following reaction: $$\mathrm{MO}(s)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{MCO}_{3}(s)$$ A \(2.85\) \(-\mathrm{g}\) sample containing only \(\mathrm{MgO}\) and \(\mathrm{CuO}\) is placed in a 3.00 -L container. The container is filled with \(\mathrm{CO}_{2}\) to a pressure of \(740 .\) torr at \(20 .^{\circ} \mathrm{C}\). After the reaction has gone to completion, the pressure inside the flask is \(390 .\) torr at \(20 .^{\circ} \mathrm{C}\). What is the mass percent of \(MgO\) in the mixture? Assume that only the \(MgO\) reacts with \(\mathrm{CO}_{2}\)

The total volume of hydrogen gas needed to fill the Hindenburg was \(2.0 \times 10^{8} \mathrm{L}\) at 1.0 atm and \(25^{\circ} \mathrm{C}\). Given that \(\Delta H_{\mathrm{f}}^{\circ}\) for \(\mathrm{H}_{2} \mathrm{O}(l)\) is \(-286 \mathrm{kJ} / \mathrm{mol},\) how much heat was evolved when the Hindenburg exploded, assuming all of the hydrogen reacted to form water?

A \(20.0\) -\(\mathrm{L}\) stainless steel container at \(25^{\circ} \mathrm{C}\) was charged with \(2.00\) atm of hydrogen gas and \(3.00\) atm of oxygen gas. A spark ignited the mixture, producing water. What is the pressure in the tank at \(25^{\circ} \mathrm{C} ?\) If the exact same experiment were performed, but the temperature was \(125^{\circ} \mathrm{C}\) instead of \(25^{\circ} \mathrm{C},\) what would be the pressure in the tank?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free