A gas is compressed at a constant pressure of 0.800atm from 9.00 L to 2.00 L. In the process,
400 J of energy leaves the gas by heat. (a) What is the work done on the gas? (b) What is the
change in its internal energy?
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Student
Student
we are required answer
To find the answers to the questions, you can use the equations related to work and internal energy in a thermodynamic process.
(a) To find the work done on the gas, you can use the equation:
Work = Pressure * Change in Volume
In this case, the pressure is constant at 0.800 atm and the change in volume is (2.00 L - 9.00 L) = -7.00 L.
Substituting these values into the equation, we get:
Work = 0.800 atm * (-7.00 L) = -5.60 atm * L
Therefore, the work done on the gas is -5.60 atm * L.
(b) To find the change in internal energy of the gas, you can use the first law of thermodynamics:
Change in Internal Energy = Heat Added - Work
In this case, the heat added is -400 J (given that 400 J of energy leaves the gas), and the work done is -5.60 atm * L (as found in part (a)).
Substituting these values into the equation, we get:
Change in Internal Energy = -400 J - (-5.60 atm * L) = -400 J + 5.60 atm * L
Therefore, the change in internal energy of the gas is -400 J + 5.60 atm * L.