For a monatomic ideal gas, the internal energy \( U \) is given by \( U = \frac{3}{2} nRT \), where \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is the temperature.
The change in internal energy \( \Delta U \) is \( \frac{3}{2} nR \Delta T \).
The heat supplied \( Q \) at constant pressure (isobaric process) is given by \( Q = n C_p \Delta T \), where \( C_p \) is the molar heat capacity at constant pressure.
For a monatomic ideal gas, \( C_p = \frac{5}{2} R \).
So, \( Q = n \left( \frac{5}{2} R \right) \Delta T = \frac{5}{2} nR \Delta T \).
The work done \( W \) by the gas during expansion at constant pressure is \( W = P \Delta V = nR \Delta T \) (from the ideal gas law \( PV = nRT \)).
Now, let's find the percentage of heat supplied used for external work:
$$ \frac{W}{Q} \times 100\
= \frac{nR \Delta T}{\frac{5}{2} nR \Delta T} \times 100\
= \frac{1}{\frac{5}{2}} \times 100\
= \frac{2}{5} \times 100\
= 40\
$$
Next, let's find the percentage of heat supplied used to increase internal energy:
$$ \frac{\Delta U}{Q} \times 100\
= \frac{\frac{3}{2} nR \Delta T}{\frac{5}{2} nR \Delta T} \times 100\
= \frac{\frac{3}{2}}{\frac{5}{2}} \times 100\
= \frac{3}{5} \times 100\
= 60\
$$
Therefore, the percentages of heat supplied used to do external work and to increase internal energy are 40
and 60
, respectively.