A) \[{{0}^{o}}C\]
B) Its critical temperature
C) Absolute zero
D) Its Boyle temperature
Correct Answer: A
Solution :
\[{{V}_{t}}={{V}_{o}}(1+{{\alpha }_{v}}t)\] \[\because ({{V}_{2}}-{{V}_{1}})=\Delta V={{V}_{o}}\alpha ({{t}_{2}}-{{t}_{1}})\] if \[{{t}_{2}}-{{t}_{1}}={{1}^{o}}\]then \[\Delta V=\alpha {{V}_{o}}\] For every \[{{1}^{o}}C\]increase in temperature, the volume of a given mass of an ideal gas increases by a definite fraction \[\frac{1}{273.15}\]of \[{{V}_{o}}\]. Here \[{{V}_{o}}\] is volume at \[{{0}^{o}}C\] temperature.
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