A) \[35{}^\circ C\]
B) \[48{}^\circ C\]
C) \[65{}^\circ C\]
D) \[14{}^\circ C\]
Correct Answer: B
Solution :
The velocity in a gas is given \[c=\sqrt{\frac{3RT}{{{V}_{\rho }}}}\]given \[{{\rho }_{{{N}_{2}}}}:{{\rho }_{{{O}_{2}}}}=14:16\] Hence, \[c\propto \sqrt{\frac{T}{\rho }}\] Since, velocity in nitrogen and oxygen is same \[\sqrt{\frac{{{T}_{{{N}_{2}}}}}{{{\rho }_{{{N}_{2}}}}}}=\sqrt{\frac{{{T}_{{{O}_{2}}}}}{{{\rho }_{{{O}_{2}}}}}}\] or \[{{T}_{{{N}_{2}}}}{{\rho }_{{{O}_{2}}}}={{T}_{{{O}_{2}}}}{{\rho }_{{{N}_{2}}}}\] \[{{T}_{{{N}_{2}}}}=\frac{{{\rho }_{{{N}_{2}}}}}{{{\rho }_{{{O}_{2}}}}}\times {{T}_{{{O}_{2}}}}=\frac{14}{16}\times 55=48.125\] \[\approx 48{{\,}^{o}}C\]You need to login to perform this action.
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