Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Two closed bulbs of equal volume (V) containing an ideal gas initially at pressure n

Two closed bulbs of equal volume (V) containing an ideal gas initially at pressure \mathrm{P_{i}} and temperature \mathrm{T_{i}} are connected through a narrow tube of negligible volume,The temperature of one of the bulbs is then raised to \mathrm{T_{2}},The final pressure \mathrm{P_{f}} is _________

Option: 1

\mathrm{ 2 p_i\left(\frac{T_1}{T_1+T_2}\right)}


Option: 2

\mathrm{2 P i\left(\frac{T_2}{T_1+T_2}\right)}


Option: 3

\mathrm{P_i\left(\frac{T_1 T_2}{T_1+T_2}\right)}


Option: 4

None


Answers (1)

best_answer

Initially,

Number of moles pf gases in each container \mathrm{=\frac{P_i V}{R T_1}}

Total number of moles of gases in both container \mathrm{=\frac{2 P_1 V}{R T_1}}

After mixing,

No. of moles of gases left is chamber\mathrm{=\frac{P_f v}{R T_1}}

No. of moles in right chamber \mathrm{=\frac{P_{f V}}{R T_2}}

\mathrm{\therefore } Total number of moles \mathrm{\frac{P_f V}{R T_1}+\frac{P_f V}{R T_2}=\frac{P_f V}{R}\left(\frac{1}{T_1}+\frac{1}{T_2}\right) }

As total number of moles remains constant.

Hence, \mathrm{\frac{2 P_i V}{R T_1}=\frac{P_f V}{R T_1}+\frac{P_f V}{R T_2} \Rightarrow P_f=2 P_i\left(\frac{T_2}{T_1+T_2}\right) }

 

Posted by

Devendra Khairwa

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

Is it difficult to get into NIT Durgapur? JEE Main cut-offs for BTech programmes
anam-khan

JEE Main Deleted Syllabus 2025: Topics to skip this year; why no optional questions?
anam-khan

NTA Exam 2025 Live: UGC NET admit card out for all exams; JEE Main session 1 hall ticket soon

Latest Question