A monoatomic ideal gas is heated at constant pressure. The percentage of total heat used in increasing the internal energy and that used for doing external work is A and B respectively. Then the ratio, $\mathrm{A}: \mathrm{B}$ is

The first law of thermodynamics states that $\Delta Q = \Delta U + \Delta W$, where $\Delta Q$ is the heat supplied, $\Delta U$ is the change in internal energy, and $\Delta W$ is the work done by the gas.

For a monoatomic ideal gas undergoing a constant pressure process, the quantities are given by $\Delta U = n C_v \Delta T$ with $C_v = \frac{3}{2} R$, $\Delta W = n R \Delta T$, and $\Delta Q = n C_p \Delta T$ with $C_p = \frac{5}{2} R$.

The percentage of heat used to increase internal energy is:

$A = \frac{\Delta U}{\Delta Q} \times 100 = \frac{n (3R/2) \Delta T}{n (5R/2) \Delta T} \times 100 = \frac{3}{5} \times 100 = 60\%$

The percentage used for external work is:

$B = \frac{\Delta W}{\Delta Q} \times 100 = \frac{n R \Delta T}{n (5R/2) \Delta T} \times 100 = \frac{2}{5} \times 100 = 40\%$

Therefore, the ratio $A : B = 60 : 40 = 3 : 2$, and the correct choice is C.