A

We can use the formula for the heat required to change the temperature of a gas at constant pressure:
Q = nC_p \Delta T
where:
- Q is the heat (calories),
- n is the number of moles (for 1 gram-molecule of oxygen, n = 1),
- C_p is the heat capacity at constant pressure (given as 0.218 cal/g.deg),
- \Delta T is the change in temperature.

To double the volume of an ideal gas at constant pressure, we can use the ideal gas law, stated as PV = nRT. At constant pressure (P), doubling the volume (V) requires doubling the number of moles of gas or increasing the temperature. Using the relationship \Delta V \propto \Delta T, if we double the volume, we have:
V_2 = 2V_1. Therefore, \Delta T is equivalent to the change in temperature when volume doubles.
This can be calculated as: \Delta T = T_final - T_initial. Since we start at 273 K (0ºC) and double the volume, the final temperature (T_final) should be 546 K. Thus, \Delta T = 546 - 273 = 273 K.
Now substituting in the heat capacity and our found temperature change:
Q = 1 \, \text{mol} \times 0.218 \, \text{cal/g.deg} \times 273 \, \text{deg} = 59.454 \, \text{cal}.
Therefore, the amount of heat required to double the volume of the gas is approximately 59.45 cal.