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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_1\left(\frac{T_1 T_2}{T_1+T_2}\right)}


Option: 4

None


Answers (1)

best_answer

Initially,
number of moles of gases in each container    \mathrm{=\frac{P_i V}{R T_1}}
Total numbers of moles of gases is both container   \mathrm{ =\frac{2 P_1 V}{R T_1}}

After mixing,
no. of moles of gases left in 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 numbers 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_f 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

manish painkra

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