Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
I read in the book the derivation of pv=nrt, they relate the change of momentum and consider time it take for 1 collision. However, if considering a cubical volume expand at the speed that is faster than the speed of the gas molecule so that the gas molecule wont collide with the wall(i don't know if this is possible or not). At that particular instant the volume may have a volume of V while the gas havent touches the wall of the container(in this case work is done by external sources in expanding the volume of the cube). At this time, is it obey pv=nRT and can take p1v1=p2v2?(cause if this can be happen and no collision occur at the wall the pressure exert on the wall is 0)?
Dear chua ,
if you are considering the gas to expand slowly than container than it actually is like as if gas is in free space as no molecule can collide to wall. it is like an expandable space. and the pv =nrt holds for that . so it shall continue to hold. why it holds in space when it is derived in a container ? because we assume that at infinity there is a container and apply the results. Now in your case no one can apply work continuously to the wall as energy is finite and so wall shall meet the molecule at infinity and so it is actually not a feasible sitiuation.
Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.
All the best.
Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.
Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar respectively : Click here to download the toolbar..
Askiitians Expert
Ajit Singh Verma IITD
dear chua kok tong,
ACTUALLY THIS IS A HYPOTHETICAL SITUATION
BUT STILL IN MY OPINION ALL YOUR SUPPOSITIONS ARE CORRECT AND THE CONDITION SEEMS TO BE TRUE...
We are all IITians and here to help you in your IIT JEE preparation. Now you can win
We are all IITians and here to help you in your IIT JEE preparation.
Now you can win
exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian. Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar :
by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian. Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar :
query on askiitians forum and become an Elite Expert League askiitian.
Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar :
Click here
to download the toolbar..
AMAN BANSAL
Anyway the P at the formula is the pressure exert by gas on the container?
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Win Gift vouchers upto Rs 500/-
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !