One mole of an ideal diatomic gas undergoes

entropy Entropy involving ideal gases Problem: Calculate the entropy change of an ideal gas that undergoes a reversible isothermal expansion from volume V 1 to V 2. Solution: • Concepts: Isothermal processes • Reasoning: For an ideal gas PV = nRT. For an isothermal process PV = constant, dU = dQ - dW = 0. dQ = dW = PdV. • Details of the calculation: dS = dQ/T = PdV/T.ΔS = (1/T) ∫ 1 2PdV = (nR) ∫ 1 2(1/V)dV = nRln(V 2/V 1). Problem: Calculate the entropy change of 1 mole of an ideal gas that undergoes an isothermal transformation from an initial state of pressure 1.5 atm and a volume of 500 cm 3 to a final state of pressure 0.90 atm. Solution: • Concepts: Isothermal processes • Reasoning: For an ideal gas PV = nRT. For an isothermal process PV = constant, dU = dQ - dW = 0, dQ = dW = PdV. • Details of the calculation: dS = dQ/T = PdV/T.ΔS = (1/T) ∫ A BPdV = (nR) ∫ A B(1/V)dV = nRln(V B/V A) = nRln(P A/P B). Here n = 1, ΔS = 8.31 * ln(1.5/0.9) J/K = 4.2 J/K. Problem: One mole of an ideal gas is compressed at 60 oC isothermally from 5 atm to20 atm. (a) Find the work done. (b) Find the entropy change for the gas and interpret its algebraic sign Gas constant : 8.314 J/(mol K) Solution: • Concepts: Ideal gas law: PV = nRT, work done on the system: W = -∫PdV Energy conservation: ΔU = ΔQ + ΔW Change in entropy:ΔS = ∫ i f dS = ∫ i f dQ r/T The subscript r denotes a reversible path. • Reasoning: Using the ideal gas law we can find the work done on the system. Using ΔU = ΔQ + ΔW we ca...

This problem has been solved! You'll get a detailed solution from a subject matter expert that helps you learn core concepts. See Answer See Answer See Answer done loading Question:130 1 Reversible Process. One mole of diatomic ideal gas undergoes a reversible carnot cycle shown in the figure. Processes 1 + 2 and 3 + 4 are reversible isothermal process while processes 2 + 3 and 4 + 1 are reversible adiabatic process. What are the en- tropy changes for each process in the cycle? (Plot is not to scale!) Pressure (kPa) 2 4 80 3 0.0130 130 1 Reversible Process. One mole of diatomic ideal gas undergoes a reversible carnot cycle shown in the figure. Processes 1 + 2 and 3 + 4 are reversible isothermal process while processes 2 + 3 and 4 + 1 are reversible adiabatic process. What are the en- tropy changes for each process in the cycle? (Plot is not to scale!) Pressure (kPa) 2 4 80 3 0.0130 0.0197 Volume (mº) Previous question Next question

The ideal gas law relates four macroscopic properties of ideal gases (pressure, volume, number of moles, and temperature). If we know the values of three of these properties, we can use the ideal gas law to solve for the fourth. In this video, we'll use the ideal gas law to solve for the number of moles (and ultimately molecules) in a sample of gas. Created by Sal Khan. Yeah mm Hg (millimeters of mercury) is a unit of pressure. It's also called a Torr after Italian scientist Evangelista Torricelli who invented the barometer. The definition itself it based on the mercury barometer which measures atmospheric pressure. The way a barometer works is if you have a thin tube closed at one end with all the air evacuated from it so that it is a vacuum inside. Placing that tube with the open air into a body liquid will result in some of the liquid flowing into the tube and reaching a certain height above the rest of the liquid's surface. Essentially what happens is that the liquid is forced into the tube by the pressure of the atmosphere pushing on the liquid's surface. The more force applied to the liquid's surface, the higher the level inside the tube rises. So what the barometer is really doing is measuring atmospheric pressure based on the height of that liquid. Now the type of liquid you choose also determines how high the liquid level rises to; specifically it depends on its density. The actual numerical height is determined by Jurin's law. Without getting too much into the ma...

This problem has been solved! You'll get a detailed solution from a subject matter expert that helps you learn core concepts. See Answer See Answer See Answer done loading Question:Question B(B.) A sample of 1.00 mole of a diatomic ideal gas is initially at temperature 265 K and volume 0.200 m3. The gas first undergoes an isobaric expansion, such that its temperature increases by 120.0 K. It then undergoes an adiabatic expansion so that its final volume is 0.360 m3.i.Sketch a PV diagram for the two-step process, including labeled Question B(B.) A sample of 1.00 mole of a diatomic ideal gas is initially at temperature 265 K and volume 0.200 m3. The gas first undergoes an isobaric expansion, such that its temperature increases by 120.0 K. It then undergoes an adiabatic expansion so that its final volume is 0.360 m3.i.Sketch a PV diagram for the two-step process, including labeled initial, final, and intermediate (after the isobaric process, before the adiabatic process) states, and a two-part curve/path with an arrow indicating direction. Label the initial state “i”, the final state “f”, and the intermediate state “b”. Write down the known values for P, T, and Vat each point, e.g. Ti = 265 K, and Tb = 385 K.3. (0.7 points) What is the initial pressure of the gas, Pi, in kPa?4. (1.1 points) What is the total heat transfer, Q, to the gas, in J? 5. (1.5 points) What is the total work done on the gas, W, in J?

Gases are complicated. They're full of billions and billions of energetic gas molecules that can collide and possibly interact with each other. Since it's hard to exactly describe a real gas, people created the concept of an Ideal gas as an approximation that helps us model and predict the behavior of real gases. The term ideal gas refers to a hypothetical gas composed of molecules which follow a few rules: If this sounds too ideal to be true, you're right. There are no gases that are exactly ideal, but there are plenty of gases that are close enough that the concept of an ideal gas is an extremely useful approximation for many situations. In fact, for temperatures near room temperature and pressures near atmospheric pressure, many of the gases we care about are very nearly ideal. If the pressure of the gas is too large (e.g. hundreds of times larger than atmospheric pressure), or the temperature is too low (e.g. − 200 C -200 \text − 2 0 0 C minus, 200, start text, space, C, end text ) there can be significant deviations from the ideal gas law. For more on non-ideal gases read Perhaps the most confusing thing about using the ideal gas law is making sure we use the right units when plugging in numbers. If you use the gas constant R = 8.31 J K ⋅ m o l R=8.31 \dfrac K kelvin K start text, k, e, l, v, i, n, space, end text, K . If you use the gas constant R = 0.082 L ⋅ a t m K ⋅ m o l R=0.082 \dfrac K kelvin K start text, k, e, l, v, i, n, space, end text, K . Units to use fo...