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At 60^{\circ} \mathrm{C} and 3 \mathrm{~atm}, a 2.5\mathrm{~L} container was filled with air. After raising the temperature to 250^{\circ} \mathrm{C}, pressure inside the container was decreased to 1 atm by opening a valve. Calculate the fraction of the total number of moles inside which were discharged, when valve was opened.

Option: 1

0.78


Option: 2

078


Option: 3

0.22


Option: 4

0.50


Answers (1)

best_answer

\mathrm{\begin{aligned} & V=2.5 \mathrm{~L} \\ & P_1=3 \mathrm{~atm} \\ & \tau_1=333 \mathrm{~K} \end{aligned}}

On having, \mathrm{V=2.5 L, \quad T_2=523 \mathrm{~K}, \quad P_2=?}

\mathrm{\begin{aligned} & \frac{P_1}{T_1}=\frac{P_2}{T_2} \\ & P_2=\frac{3 \times 523}{333}=4.71 \\ n= & \frac{P V}{R T}=\frac{3 \times 2.5}{0.0821 \times 333}=\frac{7.5}{27.339} \\ = & 0.274 \end{aligned}}

Now valve is opened till the pressure is maintained at 1 atm.

Thus, at constant V and T \mathrm{P \propto n}

\mathrm{4.71 \propto 0.274}

\mathrm{\text { so, } \quad 1 \propto n_{\text {auf }}}

\mathrm{n=\frac{0.274}{04.71}=0.058}

 Accordingly, moles which escaped out 

\mathrm{\begin{aligned} & =0.274-0.058 \\ & =0.216 \end{aligned}}

Fraction of moles that escaped

  \mathrm{\begin{aligned} & =\frac{0.216}{0.274} \\ & =0.78 \end{aligned}}

Posted by

sudhir kumar

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