Description
(a)If 375J of heat are absorbed by the system, and it has 425 cal of work done by it, compute the change in the internal energy, and interpret the sign of your result. 1 cal= 4.184J
(b)When 3.50 moles of N2(g) is heated in a piston having a constant pressure of 3.00 atm from 0.00 degrees Celsius to 225 degrees celcius, calculate the work, expressed in J, associated with this process. Interpret the sign of your answer.
Explanation & Answer
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a)Heat is absorbed by the system so q=375J
Work is done by the system ,so w=-425cal=-425*4.184 J=-1778.2 J
According to first law of thermodynamics,deltaU=q+w=-1403.2J
Interpretation of sign:Therefore, heat is taken away from the system making it exothermic and negative.
b)As the presure is constant therefore the work doe by the gas is given by the formula ,W=nR(dT)
Here dT=change in temperature=225-0=225
R=.8.314J/K
n=3.5
Wrok done=3.5*225*8.314 =6547.275J
As the pressure is constant therefore the volume of the system will increase and therefore the work is done by the system.
But work done by the system is negative therefore,work done=-6547.275 J
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