Chat with tutor

Ask Questions, Get Answers

Questions  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Thermodynamics

Two moles of a perfect gas undergoes following processes: \[\] 1. A reversible isobaric expansion from (1 atm, 20L) to (1 atm, 40L)\[\] 2. A reversible isochoric change of state from (1 atm, 40L) to (0.5 atm, 40L)\[\] 3. A reversible isothermal compression from (0.5 atm, 40L) to (1atm, 20L)\[\] Calculate the total work done involved in the above processes

$\begin {array} {1 1} (A)\;343\: J & \quad (B)\;-620.96 \: J \\ (C)\;-343\: J & \quad (D)\;620.96 \: J \end {array}$

1 Answer

Total work done $W$ is given as $W = W_1 + W_2 + W_3$
Now, $W_1 = -P \Delta V = -1 \times (40-20) = -20\: L\: -atm$
$W_2 = 0$ because $ \Delta V = 0$
$W_3 = -2.303nRT \: log \large\frac{V_2}{V_1}$$ = -2.303 \times 2 \times 0.082 \times 121.95 \times log \bigg(\large\frac{20}{40} \bigg)$
$ = 13.86\: L\: -atm$
$W = W_1 + W_2 + W_3 = -20 + 0 + 13.86 = 6.13\: L\: -atm \times 101.3 = 620.96 \: J$
Ans : (D)
Help Clay6 to be free
Clay6 needs your help to survive. We have roughly 7 lakh students visiting us monthly. We want to keep our services free and improve with prompt help and advanced solutions by adding more teachers and infrastructure.

A small donation from you will help us reach that goal faster. Talk to your parents, teachers and school and spread the word about clay6. You can pay online or send a cheque.

Thanks for your support.
Please choose your payment mode to continue
Home Ask Homework Questions
Your payment for is successful.
Clay6 tutors use Telegram* chat app to help students with their questions and doubts.
Do you have the Telegram chat app installed?
Already installed Install now
*Telegram is a chat app like WhatsApp / Facebook Messenger / Skype.