A) \[1.23\times {{10}^{5}}Pa,\,\,7.1\times {{10}^{-2}}g\]
B) \[3.21\times {{10}^{5}}Pa,\,\,7.1\times {{10}^{-2}}g\]
C) \[1.23\times {{10}^{5}}Pa,\,\,1.7\times {{10}^{-2}}g\]
D) \[3.21\times {{10}^{5}}Pa,\,\,1.7\times {{10}^{-2}}g\]
Correct Answer: A
Solution :
| let \[{{n}_{A}},{{n}_{B}}\,\,\text{and}\,\,n{{'}_{A}}\] and \[n{{'}_{B}}\] be the number of moles in bulbs A and B respectively, initially and finally respectively. |
| \[{{n}_{A}}=\frac{\left( 1atm \right)(500c{{m}^{3)}}}{\left( 237K \right)R}\] |
| And \[{{n}_{B}}=\frac{\left( 1atm \right)\left( 250c{{m}^{3}} \right)}{\left( 237K \right)R}\]\[n{{'}_{A}}=\frac{\left( Patm \right)\left( 500c{{m}^{3}} \right)}{\left( 373K \right)R}\]\[and\,n{{'}_{B}}=\frac{\left( Patm \right)\left( 250c{{m}^{3}} \right)}{\left( 237K \right)R}\] |
| Since, the total number of moles remains unchanged so, \[{{n}_{A}}+{{n}_{B}}=n{{'}_{A}}+n{{'}_{B}}\] |
| Or \[\frac{500}{273}+\frac{250}{273}=250P\left[ \frac{2}{273}+\frac{1}{273} \right]\] |
| \[or\,P=1.22\]Atmosphere=\[1.23\times {{10}^{5}}Pa\]\[\frac{n{{'}_{B}}}{{{n}_{B}}}=P=1.22\,or\,\frac{n{{'}_{B}}-{{n}_{B}}}{{{n}_{B}}}=0.22\] |
| i.e., the number of moles of air exchanged between the bulb A and B bears a ratio of 0.22 to the initial number of moles contained in the bulb B which will be the same as the mass ratio. |
| Initial mass of air in bulb B = volume\[\times \]density \[=\left( 1/4 \right)litre\times 1.29g/litre\,=32.25\times {{10}^{-2}}g\] |
| \[\therefore \] Mass of air transferred from bulb A to bulb B\[=0.22\times 32.25\times {{10}^{-2}}=7.0\times {{10}^{-2}}g\]\[=0.22\times 32.25\times {{10}^{-2}}=7.0\times {{10}^{-2}}g\] |
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