Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

A heat engine does work by using a gas at an initial pressure of 1000 Pa and vol

ID: 1389659 • Letter: A

Question

A heat engine does work by using a gas at an initial pressure of 1000 Pa and volume .1m 3. Step-by-step, it then increases the pressure to 10,000 Pa (at constant volume), increases the volume to .15m3 (at constant pressure), decreases the pressure back to 1,000 Pa (at constant volume) and returns the volume back to .1m3 (at constant pressure).

5)  How much work is done by the gas during the first step?

W = 0 J

W = 450 J

W = 500 J

W = 750 J

W = 1250 J

W = 1350 J

6)  How much work is done by the gas during the second step?

W = 0 J

W = 450 J

W = 500 J

W = 750 J

W = 1250 J

W = 1350 J

7)  How much work is done by this heat engine in one complete cycle?

W = 0 J

W = 450 J

W = 500 J

W = 750 J

W = 1250 J

W = 1350 J

8)  What is the change in internal energy during the first step?

DU = 0 J

DU = 450 J

DU = 500 J

DU = 750 J

DU = 1250 J

DU = 1350 J

9)  What is the change in internal energy during the second step?

DU = 0 J

DU = 450 J

DU = 500 J

DU = 750 J

DU = 1250 J

DU = 1350 J

10)  What is the change in internal energy over one complete cycle?

DU = 0 J

DU = 450 J

DU = 500 J

DU = 750 J

DU = 1250 J

DU = 1350 J

11)  How much heat is added to the gas in the first step?

Q = 0 J

Q = 450 J

Q = 500 J

Q = 750 J

Q = 1250 J

Q = 1350 J

12)  How much heat is added to the gas in the second step?

Q = 0 J

Q = 450 J

Q = 500 J

Q = 750 J

Q = 1250 J

Q = 1350 J

13)  How much heat is added to the gas in one complete cycle?

Q = 0 J

Q = 450 J

Q = 500 J

Q = 750 J

Q = 1250 J

Q = 1350 J

Explanation / Answer

We know that

5) W = PdV =10000(0)=0J             here change in volume =dV=0

6) W = PdV =10000(0.15-.1) =500J

7) Work in third step= 1000(0.15-1) = 50J

            Total work = 500-50 =450J

          

8)   Change in Internal energy =dE = Q-W = Q-0=Q= nCvT

    Change in internal energy here in this case will be givnen by dE =nCvT

        So there is not enough information present to deal with this problem because we dont know the type of gas exactly.

9) Not enough information , same case to section 8

10) oJ ( becasue for the whole reversible process it will be zero, it depends on the initial and final point)

11) Q=nCvdT     ( not enough information is provided)

12) Q=nCvdt       ( not enough iformation is provided for calculation)

13) Q=450J           

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote