TS EAMCET 5-Aug-2021 Shift 1 Question Paper

Given,
Initial temperature of container, Ti=300K
Initial and final velocity of particle be vi and vf
where, vf=4vi and final temperature of container be Tf
As we know that,
vav=√‌

8RT

πm

where, v‌av ‌ is average velocity, R is universal gas constant, i.e. 8.314J−K,T is temperature and m be molar mass.
On squaring both sides, we get
vav2=‌

8RT

πm

⇒‌vav2‌∝T
‌∴‌(‌

vf

vi

)2‌=(‌

Tf

Ti

)⇒(‌

4vi

vi

)2=(‌

Tf

Ti

)
⇒‌16×Ti‌=Tf⇒Tf=16×300
‌=4800K
‌∴‌Tf‌=(4800−273)∘C=4527∘C