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Standard XII

Chemistry

Ideal Gas Equation

In a 2.5 L ...

Question

In a 2.5 $L$ flask at $27^{\circ} C$ temperature, the pressure of a gas was found to be 8 $a t m$ . If $2.41 \times 10^{23}$ molecules of the same gas are introduced into the container, the temperature changed to $T_{2}$ . The pressure of gas is found to be 10 $a t m$ . Find out the value of $T_{2}$ :

253

K

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347

K

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230

K

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370

K

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Solution

The correct option is B 253 $K$
$6.023 \times 10^{23}$ molecules corresponds to 1 $m o l e$
$2.41 \times 10^{23}$ molecules corresponds to $= \frac{2.41 \times 10^{23}}{6.023 \times 10^{23}} = 0.4 m o l e s$
Case (i)
$P_{1} V_{1} = n_{1} R_{1} T_{1}$
$8 (2.5) = n_{1} (0.08) (300)$
$n_{1} = \frac{8 (2.5)}{(0.08) (300)} = 0.833 m o l e s$
Total moles $n_{2} = 0.833 + 0.4$
$n_{2} = 1.233$
Case (ii)
$P_{2} V_{2} = n_{2} R T_{2}$
$10 (2.5) = 1.233 (0.08) T_{2}$
$T_{2} = 253.4$

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In a 2.5 L flask at 27∘C temperature, the pressure of a gas was found to be 8 atm. If 2.41×1023 molecules of the same gas are introduced into the container, the temperature changed to T2. The pressure of gas is found to be 10 atm. Find out the value of T2:

The correct option is B 253 K6.023×1023 molecules corresponds to 1 mole2.41×1023 molecules corresponds to =2.41×10236.023×1023=0.4 molesCase (i)P1V1=n1R1T18(2.5)=n1(0.08)(300)n1=8(2.5)(0.08)(300)=0.833 molesTotal moles n2=0.833+0.4n2=1.233Case (ii)P2V2=n2RT210(2.5)=1.233(0.08)T2T2=253.4