The pressure in an ideal gas cut in half slowly, while being kept in a container with rigid walls. In the process, 26j kj of heat left the gas. A.how much work was done during this process? B.what is the change in internal energy of the gas during this process?


Oh sorry.
Its half slowly not halfway.
Tnx

2.If an engine, an almost ideal gas Is compressed adiabatically to half its volume. In doing so, 1850j of work is done on the gas. A. how much heat flows into or out of the gas? B. what is the change in internal energy of the gas? C. does it temp. rose or fall? :D

3. An ideal gas expands at a constant total pressure of 3.0atm from 400 ml to 660ml. Heat them flows out of the gas at constant volume, and the pressure and temp. are allowed to drop untll temp. reaches its original value . Calculate A. the total work done by the gas in the process, and B. the total heat flow into the gas.:D

1. You sure you didn't skip a word in there? I don't understand "The pressure in an ideal gas cut in halfway"

2.
A. What is the meaning of adiabatically? The answer lies in the definition.
B. Give me the first law in Thermodynamics. The answer is in the definition again.
C. What is the internal energy of a gas? Once more, the answer is in the definition itself.

3.
A. Work = P \Delta V (pressure times change in volume)

B. Use the principle of conservation of energy, or the first law of thermodynamics \Delta U = \Delta W + \Delta Q

4. Consider the following two-step process.it is allowed follow out of an ideal gas at constant volume so that it pressure drops from 2.2atm to 1.4 atm. Then the gas expands at constant pressure from vol. 6,8 to 9.3l, where the temp. reaches its original value. Calculate A. the total work done by the gas in the process. B. the change in internal energy of the gas in the process and C. the total heat flow into or out of the gas.



I don't know how to attach file.
there's an illustration given. But the given is there already.

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1. A. Use W = P \Delta V

B. Use the first law of thermodynamics, \Delta U = \Delta Q + \Delta W

4.
A. Work = P \Delta V (pressure times change in volume)

B. What is internal energy?

C. Use the first law of thermodynamics \Delta U = \Delta W + \Delta Q You have the change in internal energy from B, the work done from A and you need Q, the heat involved. If it's negative, that means more heat was absorbed, and you can guess what happens in the opposite case.

For a change, post the answers you've got in all the different questions you've posted.