A gas experiences isobaric expansion at 2.0 atm from 4.2 m3 to 7.5 m3 . How...
A gas experiences isobaric expansion at 2.0 atm from 4.2 m3 to 7.5 m3 . How much work was done by the gas on the environment?
A room in a spaceship experiences catastrophic failure and loses its atmosphere to space. The room, which is 8m x 8m x 8m, contained 54682 mols of air at standard temperature and pressure. How much work does the atmosphere do if the venting of atmosphere is adiabatic and the final temperature of the gas is 8.21 K?
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![W=PdV = \left [202650\cdot (7.5-4.2) \right ]\: J= 668745\: J](http://img.homeworklib.com/questions/801dbc30-c534-11ea-b0d6-d3dbf80f52d3.png?x-oss-process=image/resize,w_560)
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A gas experiences isobaric expansion at 2.0 atm from 4.2
m3 to 7.5 m3 . How...
12. 1 mole of an ideal gas undergoes an isothermal expansion from V1 = 1.4L followed...
12. 1 mole of an ideal gas undergoes an isothermal expansion from V1 = 1.4L followed by isobaric compression, p = cst.if P1 = 4.4atm, p2 = 1.7atm → ?- m calculate the work done by gas during the expansion. Express work in J = N·m! • For isothermal processes, AT = 0 T = cst → w=faw=fr&v=/MRT AV 594 Show your work like: `x-int_0^5 v(t)dt rarr x-int_0^5(-4*t)dt=-50 m 13. 1 mole of an ideal gas undergoes an isothermal expansion...
28. The work done by an ideal gas in an isothermal expansion from v volume Vs...
28. The work done by an ideal gas in an isothermal expansion from v volume Vs to volume V, is glven by the formula: w = nRT In(V/V) spheric pressure (1 atm) is 101.3 KPa. If 1.0 L of He gas at room temperature 0 atm of (208C) and 1. work is done on the gas? pressure is compressed isothermally to a volume of 100 mL, how much 02 E 23x1 D) 2.3 x 10 A) 5.6 ku B) 4.7...
During an adiabatic expansion the temperature of 0.490 mol of argon (Ar) drops from 64.0 °C...
During an adiabatic expansion the temperature of 0.490 mol of argon (Ar) drops from 64.0 °C to 10.0 °C. The argon may be treated as an ideal gas. How much work does the gas do? Express your answer with the appropriate units. t μΑ ? 288 W J Submit Previous Answers Request Answer X Incorrect; Try Again; 2 attempts remaining You may have forgotten that the process is adiabatic, not isobaric. Recall the definition of this process, and use the...
1. What is the purpose of a constant volume gas thermometer. How does it work? 2. What is the difference, if any, betwee...
1. What is the purpose of a constant volume gas thermometer. How does it work? 2. What is the difference, if any, between a isothermal and adiabatic thermodynamic process. Draw them on a pV diagram. 3. Does the First Law of Thermodynamics apply to solids and liquids? Why or why not. 4. What is the most important heat transfer mechanism on earth where there is an atmosphere? Explain how this can cool a nuclear or coal fired power plant, and...
12. 1 mole of an ideal gas undergoes an isothermal expansion from V1 = 1.4L followed by isobaric compression, p = cst.if P1 = 4.4atm, p2 = 1.7atm → ?- m calculate the work done by gas during the expansion. Express work in J = N·m! • For isothermal processes, AT = 0 T = cst → w=faw=fr&v=/MRT AV 594 Show your work like: `x-int_0^5 v(t)dt rarr x-int_0^5(-4*t)dt=-50 m 13. 1 mole of an ideal gas undergoes an isothermal expansion...
28. The work done by an ideal gas in an isothermal expansion from v volume Vs to volume V, is glven by the formula: w = nRT In(V/V) spheric pressure (1 atm) is 101.3 KPa. If 1.0 L of He gas at room temperature 0 atm of (208C) and 1. work is done on the gas? pressure is compressed isothermally to a volume of 100 mL, how much 02 E 23x1 D) 2.3 x 10 A) 5.6 ku B) 4.7...
During an adiabatic expansion the temperature of 0.490 mol of argon (Ar) drops from 64.0 °C to 10.0 °C. The argon may be treated as an ideal gas. How much work does the gas do? Express your answer with the appropriate units. t μΑ ? 288 W J Submit Previous Answers Request Answer X Incorrect; Try Again; 2 attempts remaining You may have forgotten that the process is adiabatic, not isobaric. Recall the definition of this process, and use the...
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