A) 2
B) 1
C) \[\frac{1}{2}\]
D) \[\frac{1}{3}\]
Correct Answer: A
Solution :
[a] Given conditions \[{{V}_{1}}=16.4L,{{V}_{2}}=5L\] \[{{P}_{1}}=1.5\,atm,\,\,{{P}_{2}}=4.1atm\] \[{{T}_{1}}=273+27=300K,\] \[{{T}_{2}}=273+227=500K\] Applying gas equation, \[\frac{{{P}_{1}}{{V}_{1}}}{{{P}_{2}}{{V}_{2}}}=\frac{{{n}_{1}}{{T}_{1}}}{{{n}_{2}}{{T}_{2}}}\] \[\frac{{{n}_{1}}}{{{n}_{2}}}=\frac{{{P}_{1}}{{V}_{1}}{{T}_{1}}}{{{P}_{2}}{{V}_{2}}{{T}_{2}}}\] \[\therefore \frac{1.5\times 16.4\times 500}{4.1\times 5\times 300}=\frac{2}{1}\]You need to login to perform this action.
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