Q.
1/2 mole of helium gas is contained in a container at S.T.P. The heat energy (in Joule) needed to double the pressure of the gas, keeping the volume constant (heat capacity of the gas = 3Jg-1 K-1) is
see full answer
Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!
Real Strategies. Real People. Real Success Stories - Just 1 call away
An Intiative by Sri Chaitanya
answer is 1638.
(Unlock A.I Detailed Solution for FREE)
Ready to Test Your Skills?
Check your Performance Today with our Free Mock Test used by Toppers!
Take Free Test
Detailed Solution
Here, n=12 ,Heat capacity=Cv=3Jg−1K−1,M=4gmol−1∴Cv=Mcv=4×3=12Jmol−1K−1 At constant volume P∝T∴P2P1=T2T1=2⇒T2=2T1 Rise in temperature ΔT=T2−T1=2T1−T1=T1=273K Heat required, ΔQ=nCvΔT=12×12×273=1638J
Watch 3-min video & get full concept clarity