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