A monatomic ideal gas expands slowly to twice its original volume, doing 340 JJ of work in the process. Part APart complete Find the heat added to the gas if the process is isothermal. QQ = 340 JJ SubmitPrevious Answers Correct Important: If you use this answer in later parts, use the full unrounded value in your calculations. Part B Find the change in internal energy of the gas if the process is isothermal. ΔUΔU = 340 JJ SubmitPrevious AnswersRequest Answer Incorrect; Try Again; 4 attempts remaining Part C Find the heat added to the gas if the process is adiabatic. QQ = nothing JJ SubmitRequest Answer Part D Find the change in internal energy of the gas if the process is adiabatic. ΔUΔU = nothing JJ SubmitRequest Answer Part E Find the heat added to the gas if the process is isobaric. QQ = nothing JJ SubmitRequest Answer Part F Find the change in internal energy of the gas if the process is isobaric. ΔUΔU = nothing JJ

A) if the system is isothermal then all the heat added to the system will be used to do work (since none is used to raise the temperature of the gas). The heat added will be equal to the work done = 340 J

B) change in internal energy of the system of the process is isothermal will be zero, since there is no rise in temperature.

C) an adiabatic process is one involving no heat loss or gain through the system, Therefore heat gain will be zero

D) if the process is adiabatic then there is no heat loss or gain through the system and hence there is no change in temperature. Change in internal energy will be zero

E) if the process is isobaric then, there is no work done and the total heat to the system is equal zero

F) if there is no work done, and no heat added, then the internal energy will be equal zero.

? you expect us to answer this? ? this is your schoolwork, i don't think we can with this, i think it would considering being

frequency, z, and the collision density, z, in carbon dioxide (co2), r 180 pm at 30 °c and 120 kpa