Final answer:
The work done by an ideal monoatomic gas expanding at constant pressure is (2/5)q since a fraction of the heat added changes the internal energy and the rest does the work. For a monoatomic gas, as the specific heat at constant volume (Cv) relates to the heat added (q), option (c) correctly represents the work done as (2/5)q.
Step-by-step explanation:
When a vessel contains an ideal monoatomic gas that expands at constant pressure, the work done by the gas during expansion can be calculated using principles of thermodynamics. According to the first law of thermodynamics, heat added to a system, q, equals the change in internal energy, ΔU, plus the work done by the system, W. For an ideal monoatomic gas undergoing an isobaric (constant pressure) expansion, the change in internal energy (ΔU) is the increase in temperature times the number of moles and the specific heat capacity at constant volume, Cv.
Since we're dealing with a monoatomic ideal gas, the relationship between the specific heats at constant pressure, Cp, and at constant volume, Cv, is given by Cp = Cv + R, where R is the ideal gas constant. For a monoatomic gas, Cv is equal to (3/2)R, thus Cp is (5/2)R. Applying the first law of thermodynamics, q = ΔU + W, and for a constant pressure process, ΔU = (3/2)nRΔT. Consequently, the work done by the gas W is q - ΔU = q - (3/5)q = (2/5)q, where (3/5)q comes from the relation ΔU to q for a monoatomic gas (Cv/Cp = 3/5). Therefore, the work done in expansion by the ideal monoatomic gas at constant pressure is (2/5)q, which corresponds to option (c).