Pressure remaining the same, the volume of a given mass of an ideal gas increases for every degree centigrade rise in temperature by definite fraction of its volume at:
Correct option is A. $$0^{o} C$$
$$V_{t} = V_{o}\ (1+\alpha_{v} t)$$
$$\because (V_{2} - V_{1}) = \Delta V = V_{0} \alpha (t_{2} - t_{1})$$
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 $$ \dfrac{1}{273.15}$$ of $$V_{o}$$.
Here $$V_{o}$$ is volume at $$0^{o}C$$ temperature.