Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A gas enclosed in a cylinder by a piston that can move without friction is warmed, and 1000 J of heat enters the gas. Assuming that the volume of the gas is constant, the change in the internal energy of the gas is a) \(0 .\) b) 1000 J. c) -1000 J. d) none of the above.

Short Answer

Expert verified
Answer: The change in internal energy of the gas is 1000 J.

Step by step solution

01

Determine the work done by the system

Since the volume of the gas is constant, there is no expansion or contraction of the gas, and consequently, the work done by the system (W) is zero.
02

Apply the first law of thermodynamics

Now we can apply the first law of thermodynamics, which is given by: ∆U = Q - W In this case, Q = 1000 J (heat added to the system) and W = 0 J (no work done by the system). Then, plug in the values to the equation: ∆U = 1000 J - 0 J
03

Calculate the change in internal energy

Now, we can calculate the change in internal energy of the gas: ∆U = 1000 J The change in internal energy of the gas is 1000 J, which corresponds to answer choice (b).

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

In which type of process is no work done on a gas? a) isothermal b) isochoric c) isobaric d) none of the above

A thermos bottle fitted with a piston is filled with a gas. Since the thermos bottle is well insulated, no heat can enter or leave it. The piston is pushed in, compressing the gas. a) What happens to the pressure of the gas? Does it increase, decrease, or stay the same? b) What happens to the temperature of the gas? Does it increase, decrease, or stay the same? c) Do any other properties of the gas change?

Suppose \(0.0100 \mathrm{~kg}\) of steam (at \(100.00^{\circ} \mathrm{C}\) ) is added to \(0.100 \mathrm{~kg}\) of water (initially at \(19.0^{\circ} \mathrm{C}\) ). The water is inside an aluminum cup of mass \(35.0 \mathrm{~g} .\) The cup is inside a perfectly insulated calorimetry container that prevents heat exchange with the outside environment. Find the final temperature of the water after equilibrium is reached.

Why does tile feel so much colder to your feet after a bath than a bath rug? Why is this effect more striking when your feet are cold?

For a class demonstration, your physics instructor pours \(1.00 \mathrm{~kg}\) of steam at \(100.0^{\circ} \mathrm{C}\) over \(4.00 \mathrm{~kg}\) of ice at \(0.00^{\circ} \mathrm{C}\) and waits for the system to reach equilibrium, when he will measure the temperature. While the system reaches equilibrium, you are given the latent heats of ice and steam and the specific heat of water: \(L_{\text {ice }}=3.33 \cdot 10^{5} \mathrm{~J} / \mathrm{kg}, L_{\text {steam }}=2.26 \cdot 10^{6} \mathrm{~J} / \mathrm{kg}\), \(c_{\text {water }}=4186 \mathrm{~J} /\left(\mathrm{kg}^{\circ} \mathrm{C}\right) .\) You are asked to calculate the final equilibrium temperature of the system. What value do you find?

See all solutions

Recommended explanations on Physics 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