Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
Close
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

⇐ BackProceed to Checkout ⇒
There are no items in this cart.
Continue Shopping

One mole of a diatomic gas is heated under “ Kibolinsky Process” in which gas pressure is temperture controlled according to law P=4/5*CT^(3/2) . Where C is Kibolinsky constant. In this process, gas is heated by 300 K temperature.Answer the following question: The molar specific heat of the gas?

One mole of a diatomic gas is heated under “ Kibolinsky Process” in which gas
pressure is temperture controlled according to law P=4/5*CT^(3/2) . Where C is Kibolinsky constant. In this process, gas is heated by 300 K temperature.Answer
the following question:
The molar specific heat of the gas?

Grade:11

1 Answers

Debojyoti Chaki
11 Points
4 years ago
Answer is 2RHow?Sol.P=4/5CT^3/2......(1)(PV)^3/2=(nRT)^3/2.......(2)(1)/(2)= P(V^3/2)=const........n=3C=R/gama-1+R/1-nPutting the values we get C=2R.

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

Other Related Questions on Thermal Physics

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Register ICAT

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More