33 Views

an ideal gas is enclosed in a cylinder at pressure of 2atm and temperature,300k the mean time between two successive collisions is 6×10^-8.If the pressure is doubled and temperature is increased to 500k,the mean time between two successive collisions will be close to?


ismailmoin09 28th Jun, 2021
Answer (1)
ginnisachdeva02_9364270 Student Expert 28th Jun, 2021

Hi Aspirant!

As we know that,

t is directly proportional to (Volume/Velocity)

=> Volume is directly proportional to (T/P)

Therefore, t is directly proportional to [(sqr root T)/ P ]

Now, t1 / (6x10^-8) = [(sqr root 500)/2P ] * [P / (sqr root 300) ]

=> t1 = 3.8x10^-8

=> t1 ~= 4x10^-8 seconds

Hope it helps to clear your dilemma, thankyou!

Related Questions

Amity University | M.Tech Adm...
Apply
Ranked amongst top 3% universities globally (QS Rankings).
Amity University Noida MBA Ad...
Apply
Amongst top 3% universities globally (QS Rankings)
Graphic Era (Deemed to be Uni...
Apply
NAAC A+ Grade | Among top 100 universities of India (NIRF 2024) | 40 crore+ scholarships distributed
Amity University Noida B.Tech...
Apply
Among Top 30 National Universities for Engineering (NIRF 2024) | 30+ Specializations | AI Powered Learning & State-of-the-Art Facilities
UPES MBA Admissions 2025
Apply
Ranked #41 amongst institutions in Management by NIRF | 100% Placement | Last Date to Apply: 15th July
Amity University, Noida BBA A...
Apply
Ranked amongst top 3% universities globally (QS Rankings)
View All Application Forms

Download the Careers360 App on your Android phone

Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile

150M+ Students
30,000+ Colleges
500+ Exams
1500+ E-books