A

To determine the amount of heat required to double the volume of 1 gram-molecule (molar mass = 32 g) of oxygen gas at constant pressure, we can use the formula:

$Q = n C_p \Delta T$
Where:
- $Q$ = Heat added (calories)
- $n$ = Number of moles of gas (1 gram-molecule = 1 mole)
- $C_p$ = Heat capacity at constant pressure (given as 0.218 cal/g.deg)
- $\Delta T$ = Change in temperature in degrees Celsius.

Since we are considering an ideal gas, we can use the ideal gas law to relate the volume and temperature change at constant pressure. According to the ideal gas law:

$PV = nRT$

For a constant pressure process, if the volume is doubled, the final temperature can be determined by:

$V_1 / T_1 = V_2 / T_2 \implies 2V_1 / T_2\implies T_2 = 2T_1$

Now substituting $T_1 = 273 K$ (0ºC) into the equation, we get:

$T_2 = 2 * 273 K = 546 K$

The change in temperature is:

$\Delta T = T_2 - T_1 = 546 K - 273 K = 273 K$

Next, since we have 32 g of oxygen:

Here, $n = 1$ mole (32g of O2).

The heat required (Q) can thus be calculated as:

$Q = n C_p \Delta T = 1 \times 0.218 \text{ cal/g.deg} \times 273 \text{ deg} = 59.454 \text{ cal}$

Therefore, the amount of heat that should be imparted to the gas to double its volume is approximately 59.454 calories.