A) 327 K
B) \[327{}^\circ C\]
C) \[54{}^\circ C\]
D) 150K
Correct Answer: B
Solution :
For an ideal gas, at constant pressure \[V\propto T\] \[\therefore \] \[\frac{{{V}_{2}}}{{{V}_{1}}}=\frac{{{T}_{2}}}{{{T}_{1}}}\Rightarrow {{T}_{2}}=\frac{{{V}_{2}}}{{{V}_{1}}}{{T}_{1}}\] Given: \[{{V}_{1}}={{V}_{0,}}{{V}_{2}}=2{{V}_{0}}\] \[{{T}_{1}}=27{}^\circ C=300K\] \[{{T}_{2}}=\frac{2{{V}_{0}}}{{{V}_{0}}}\times 300=600K\] \[=(600-273K{}^\circ C=327{}^\circ C\]
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