Email
Chat with tutor
logo

Ask Questions, Get Answers

X
 
Questions  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Thermodynamics
Answer
Comment
Share
Q)

The value of $ \Delta E$ for reversible isothermal evaporation of $90g$ of water at $100^{\circ}C$. Assume that water vapor behaves as ideal gas and $ \Delta H_{vap}$ of water is $540\: cal/g(R = 2\: cal/molK)$

$\begin {array} {1 1} (A)\;44870 \: cal & \quad (B)\;-44870\: cal \\ (C)\;52330 \: cal & \quad (D)\;-52330\: cal \end {array}$

1 Answer

Comment
A)
The total heat change, $ \Delta H = 90 \times 540 = 48600 \: cal$
Now, $ \Delta H = \Delta E + P\Delta V$
$\Delta V = (V_{vapor} – V_{liquid}) = V_{vapor} (V_{liquid} is negligible)$
$\Delta H = \Delta E + PV_{vapor} = \Delta E + nRT$
$\Rightarrow \Delta E = \Delta H – nRT$
$ = 48600 – 90/18 \times 2 \times 373 = 48600 – 3730 = 44870 \: cal$
Ans : (A)
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.
Continue
Please choose your payment mode to continue
Home Ask Homework Questions
Your payment for is successful.
Continue
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.
...