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(5%) Problem 10: A monatomic ideal gas initially fills a V0 = 0.15 m³ container at P0 = 55 kPa. The gas undergoes an isobaric expansion to V1 = 0.8 m³. Next it undergoes an isovolumetric cooling to its initial temperature T0. Finally it undergoes an isothermal compression to its initial pressure and volume. 10% Part (a) Identify the P-V diagram that correctly represents this three step cycle. 10% Part (b) Calculate the work done by the gas, W1, in kilojoules, during the isobaric expansion (first process). 10% Part (c) Calculate the heat absorbed Q1, in kilojoules, during the isobaric expansion (first process). 10% Part (d) Write an expression for the change in internal energy, ?U1 during the isobaric expansion (first process). 10% Part (e) Calculate the work done by the gas, W2, in kilojoules, during the isovolumetric cooling (second process). 10% Part (f) Calculate the heat absorbed Q2, in kilojoules, during the isovolumetric cooling (second process). 10% Part (g) Calculate the change in internal energy by the gas, ?U2, in kilojoules, during the isovolumetric cooling (second process).

          (5%) Problem 10: A monatomic ideal gas initially fills a V0 = 0.15 m³ container at P0 = 55 kPa. The gas undergoes an isobaric expansion to V1 = 0.8 m³. Next it undergoes an isovolumetric cooling to its initial temperature T0. Finally it undergoes an isothermal compression to its initial pressure and volume. 10% Part (a) Identify the P-V diagram that correctly represents this three step cycle. 10% Part (b) Calculate the work done by the gas, W1, in kilojoules, during the isobaric expansion (first process). 10% Part (c) Calculate the heat absorbed Q1, in kilojoules, during the isobaric expansion (first process). 10% Part (d) Write an expression for the change in internal energy, ?U1 during the isobaric expansion (first process). 10% Part (e) Calculate the work done by the gas, W2, in kilojoules, during the isovolumetric cooling (second process). 10% Part (f) Calculate the heat absorbed Q2, in kilojoules, during the isovolumetric cooling (second process). 10% Part (g) Calculate the change in internal energy by the gas, ?U2, in kilojoules, during the isovolumetric cooling (second process).

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(5%) Problem 10: A monatomic ideal gas initially fills a V0 = 0.15 m³ container at P0 = 55 kPa. The gas undergoes an isobaric expansion to V1 = 0.8 m³. Next it undergoes an isovolumetric cooling to its initial temperature T0. Finally it undergoes an isothermal compression to its initial pressure and volume. 10% Part (a) Identify the P-V diagram that correctly represents this three step cycle. 10% Part (b) Calculate the work done by the gas, W1, in kilojoules, during the isobaric expansion (first process). 10% Part (c) Calculate the heat absorbed Q1, in kilojoules, during the isobaric expansion (first process). 10% Part (d) Write an expression for the change in internal energy, ?U1 during the isobaric expansion (first process). 10% Part (e) Calculate the work done by the gas, W2, in kilojoules, during the isovolumetric cooling (second process). 10% Part (f) Calculate the heat absorbed Q2, in kilojoules, during the isovolumetric cooling (second process). 10% Part (g) Calculate the change in internal energy by the gas, ?U2, in kilojoules, during the isovolumetric cooling (second process).

Added by Miguel -Ngel W.

University Physics with Modern Physics

Hugh D. Young 14th Edition

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Solved by Expert Animesh Raj

12/09/2021

Step 1

The isobaric expansion would be a horizontal line moving to the right (as volume increases at constant pressure), the isovolumetric cooling would be a vertical line moving downwards (as pressure decreases at constant volume), and the isothermal compression would Show more…

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Problem 10: A monatomic ideal gas initially fills a V0 = 0.15 m³ container at P0 = 55 kPa. The gas undergoes an isobaric expansion to V1 = 0.8 m³. Next, it undergoes an isovolumetric cooling to its initial temperature T0. Finally, it undergoes an isothermal compression to its initial pressure and volume. Part (a) Identify the P-V diagram that correctly represents this three-step cycle. Part (b) Calculate the work done by the gas, W1, in kilojoules, during the isobaric expansion (first process). Part (c) Calculate the heat absorbed Q1, in kilojoules, during the isobaric expansion (first process). Part (d) Write an expression for the change in internal energy, ΔU1, during the isobaric expansion (first process). Part (e) Calculate the work done by the gas, W2, in kilojoules, during the isovolumetric cooling (second process). Part (f) Calculate the heat absorbed Q2, in kilojoules, during the isovolumetric cooling (second process). Part (g) Calculate the change in internal energy by the gas, ΔU2, in kilojoules, during the isovolumetric cooling (second process).

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00:01 So in this question the initial volume the initial volume of the monotomic ideal gas is given as v0 equals to 0 .15 meter cube the final volume is given as 0 .0 .8 meter cube the initial value of pressure is equal to 55 kios so cp will be equal to 5 r by 2 and cv will be equal to 3 r by 2 now the t0 t0 is the temperature at temperature at state 0 similarly t1 is the temperature at state 1 and t2 is the temperature at state 2 now the heat that the heat that is absorbed for the first subpart for the first subpart we have to identify the pv diagram so this will be the pv diagram this is the p axis this is the v axis this is the first process then the second process and this will be the third process now for subpart v we have to find the work done by the gas so work done by the gas work done by the gas is equal to p delta v that is equal to pressure is equal to 55 kilo pascal so 55 into 10 to the part 3 and will take the difference of the volumes so 0 .1 minus 0 .8, 0 .15 minus 0 .8.

02:09 This will be in jules.

02:10 Let's calculate the value.

02:12 55 into 10 to the power 3.

02:16 55 into 10 to the power 3 into 0 .15 minus 0 .8 gives us the value that is equal to minus 35 .75 kilojoules.

02:34 I'm sorry, this will be positive.

02:36 This will be positive because this is 0 .1 .0 .8 minus 0 .15.

02:44 This is 0 .8 minus 0 .15.

02:47 So the work done by the gas will be positive.

02:52 Now the heat absorbed.

02:55 The heat absorbed by the gas will be equal to ncp t1 minus t0.

03:03 And this is equal to n into 5 r by 2 into t1 minus t0.

03:11 This will be equal to 5 by 2 into nr t1 minus nr t0.

03:18 And that is equal to p .0v1 minus p .0v0...

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