A) \[\sqrt{2}T\]
B) T
C) 4T
D) 2T
Correct Answer: A
Solution :
For an ideal gas, the relation is \[PV=RT\] ?(i) Given \[V{{P}^{2}}=K\] ?(ii) Squaring equation (i) we get \[{{P}^{2}}{{V}^{2}}={{R}^{2}}{{T}^{2}}\] ?(iii) Now dividing equation (ii) by (iii) \[\frac{1}{V}=\frac{K}{{{R}^{2}}{{T}^{2}}}\] If volume V expands to volume 2V so \[{{T}^{2}}\propto V\] Hence \[\frac{T_{1}^{2}}{T_{2}^{2}}=\frac{V}{2V}=\frac{1}{2}\] so, \[{{T}_{2}}=\sqrt{2}\,{{T}_{1}}\]You need to login to perform this action.
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