A) \[{{P}_{2}}=P,\,\,{{T}_{2}}={{T}_{1}}\]
B) \[{{P}_{2}}={{P}_{1}},\,{{T}_{2}}=\frac{{{T}_{1}}}{2}\]
C) \[{{P}_{2}}=2{{P}_{1}},\,{{T}_{2}}={{T}_{1}}\]
D) \[{{P}_{2}}=2{{P}_{1}},\,{{T}_{2}}=\frac{{{T}_{1}}}{2}\]
Correct Answer: B
Solution :
| [b] Kinetic energy of N molecule of gas, \[E=\frac{3}{2}NkT\] |
| Initially, \[{{E}_{1}}=\frac{3}{2}{{N}_{1}}k{{T}_{1}}\] and finally, \[{{E}_{2}}=\frac{3}{2}{{N}_{2}}k{{T}_{2}}\] |
| But according to \[{{E}_{1}}={{R}_{2}}\] and \[{{N}_{2}}=2{{N}_{1}},\] |
| \[\frac{3}{2}{{N}_{1}}k{{T}_{1}}=\frac{3}{2}\,(2{{N}_{1}})k{{T}_{2}}\Rightarrow \,{{T}_{2}}=\frac{{{T}_{1}}}{2}\] |
| Since the kinetic energy constant, |
| \[\frac{3}{2}{{N}_{1}}k{{T}_{1}}=\frac{3}{2}{{N}_{2}}k{{T}_{2}}\] |
| \[\Rightarrow \] \[{{N}_{1}}{{T}_{1}}={{N}_{2}}{{T}_{2}}\] |
| \[\therefore \] \[NT=\]Constant |
| From ideal gas equation of N molecules, |
| \[PV=NkT\] |
| \[\Rightarrow \] \[{{P}_{1}}{{V}_{1}}={{P}_{2}}{{V}_{2}}\Rightarrow \,{{P}_{1}}={{P}_{2}}\] |
| [as \[{{V}_{1}}={{V}_{2}}\] and \[NT=\] constant] |
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