A) \[N\,\,molecules\,\,of\,\,{{O}_{2}}\]
B) \[2/N\,\,molecules\,of\,\,{{O}_{2}}\]
C) \[N/2\,\,molecules\,of\,\,{{O}_{2}}\]
D) \[N/4\,\,molecules\,of\,\,{{O}_{2}}\]
Correct Answer: A
Solution :
\[KE\,\,=\,\,\frac{3}{2}RT;\,\,T=-123+273=+150\,\,K\] \[\frac{3}{2}\times R\times 150=\frac{3}{2}\times 8.314\times 75=x\,J\] \[= 225 \times 8.314 =x\,J\] \[\operatorname{At}\,\,27{}^\circ C=27+223=300K\] \[KE\,\,joules\,\,=2x\,\,joules = \frac{3}{2}\,\,\times \, 8.314 \times 300\] N molecules \[\therefore ~~x joules =3\times 8.314\times 75\] In both the cases x joules correspond to N molecules.You need to login to perform this action.
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