Question

Problem zu a constant-volume process, the temperature of a perfect gas changes from T to 1;. The change in entropy of the gas

0 0
Add a comment Improve this question Transcribed image text
Answer #1

363-002 As we know, of Lue cont] OS = DO or ds = d d 50 and we also know the amount of heat added in constant volume process.as you can see that change in entropy is proportional to ln(T2/T1) but in the option it is misprint so this is the right answer.

Add a comment
Know the answer?
Add Answer to:
Problem zu a constant-volume process, the temperature of a perfect gas changes from T to 1;....
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • Write expressions in terms of the temperature T1, T2, T3, and T4 for the entropy change in each s...

    Write expressions in terms of the temperature T1, T2, T3, and T4 for the entropy change in each step of the following process for an ideal monatomic gas: a) Reversible adiabatic expansion from T1P1V to T2P2V' b) Irreversible heating at constant volume from T2P2V' to T3P3V' c) Reversible adiabatic compression from T3P3V' to T4P4V d) Irreversible cooling at constant volume from T4P4V to T1P1V. Prove that the total entropy change for the cycle is 0. Please provide explanation, work, and...

  • I'm having trouble understanding the adiabatic expansion of a perfect gas. My book gives the following...

    I'm having trouble understanding the adiabatic expansion of a perfect gas. My book gives the following graph, but I don't understand how the change in internal energy is 0 for process 1 (going from Ti,Vi to Ti,Vf). Youre expanding the gas-- would that not require work, ie: loss of internal energy? It's not like heat can be added to balance this out and keep the internal energy constant, since this is adiabatic. Why does only the temperature change (process 2)...

  • A monatomic ideal gas is initially at volume, pressure, temperature (Vi, Pi, Ti). Consider two different...

    A monatomic ideal gas is initially at volume, pressure, temperature (Vi, Pi, Ti). Consider two different paths for expansion. Path 1: The gas expands quasistatically and isothermally to (Va, Pz. T2) Path 2: First the gas expands quasistatically and adiabatically (V2, P.,T-),where you will calculate P T. Then the gas is heated quasistically at constant volume to (Va. P2 T1). a. Sketch both paths on a P-V diagram. b. Calculate the entropy change of the system along all three segments...

  • Initially, at a temperature T, and a molar volume vi, a van der Waals gas undergoes...

    Initially, at a temperature T, and a molar volume vi, a van der Waals gas undergoes a change of state to the final temperature T2 and the molar volume V2. The van der Waals gas is characterized by the two parameters a and b (cf. Eq. (3.3)). a. Show that the change in molar entropy is As = c, In 72 + R In º2 = (3.62) 01 - 6 b. A volume of 1 dm is partitioned by a...

  • Clear handwriting Ideal gas (n 2.053 mol) is heated at constant volume from ti 124.00°C to...

    Clear handwriting Ideal gas (n 2.053 mol) is heated at constant volume from ti 124.00°C to final temperature t = 244.00°C. Calculate the work and heat for the process and the change of entropy of the gas. The isobaric heat capacity of the gas is Cp,m = 28.609 J-K1-mol* Ideal gas (n 2.053 mol) is heated at constant volume from ti 124.00°C to final temperature t = 244.00°C. Calculate the work and heat for the process and the change of...

  • For a certain amount of gas at constant temperature, the pressure and volume have an inverse...

    For a certain amount of gas at constant temperature, the pressure and volume have an inverse relationship (Figure 1). This is called Boyle's law: P1V1=P2V2 Part A: A sample of ideal gas at room temperature occupies a volume of 24.0 L at a pressure of 902 torr . If the pressure changes to 4510 torr , with no change in the temperature or moles of gas, what is the new volume, V2? Express your answer with the appropriate units. Part...

  • deal gases obey the equation PV nRT, where P is the pressure of the gas, V is its volume, n is the number of moles of gas, T is its temperature, and the constant R-8.314 KPa-liters-mol-1 kelvin-1...

    deal gases obey the equation PV nRT, where P is the pressure of the gas, V is its volume, n is the number of moles of gas, T is its temperature, and the constant R-8.314 KPa-liters-mol-1 kelvin-1 (a) Find the exac t change in volume of O, gas as the pressure increases from 12.00 to 12.01 KPa, the temperature decreases from 300.0 to 299.9 degrees kelvin, and the number of moles of 0, gas changes from 1.03 to 1.01 moles....

  • For a certain amount of gas at constant temperature, the pressure and volume have an inverse...

    For a certain amount of gas at constant temperature, the pressure and volume have an inverse relationship (Figure 1). This is called Boyle's law: P1V1=P2V2 A. A sample of ideal gas at room temperature occupies a volume of 33.0 L at a pressure of 862 torr . If the pressure changes to 4310 torr , with no change in the temperature or moles of gas, what is the new volume, V2? B. If the volume of the original sample in...

  • An ideal gas at temperature T is initially confined to 1/6th of volume V of an...

    An ideal gas at temperature T is initially confined to 1/6th of volume V of an insulated gas jar. After a while the gas fills the entire jar uniformly as shown in the figure below: (a) Describe the initial state of the gas in {T, V } coordinates. (b) Describe the final state of the gas in {T, V } coordinates. (c) Calculate the change in the entropy of the gas as it goes from the initial state to the...

  • Charles’s law states that the volume of a gas is directly related to the absolute temperature...

    Charles’s law states that the volume of a gas is directly related to the absolute temperature when there is no change in the pressure or amount of gas: (Figure 1) V1T1=V2T2 Part A A sample of gas in a balloon has an initial temperature of 29 ∘C and a volume of 1020 L . If the temperature changes to 98 ∘C , and there is no change of pressure or amount of gas, what is the new volume, V2, of...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT