To determine the heat supplied at constant volume for a monoatomic gas, let's start by defining some fundamental relationships and properties of monoatomic gases.

The ratio of specific heats $$\gamma$$ for a monoatomic gas is given as:

$$\gamma = \frac{C_p}{C_v} = \frac{5}{3}$$

Where:

$$C_p$$: Specific heat at constant pressure
$$C_v$$: Specific heat at constant volume

The work done at constant pressure, $$\mathrm{W}$$, for a rise in temperature $$\Delta T$$ is defined as:

$$W = P \Delta V$$

Using the ideal gas law, $$PV = nRT$$, we know that at constant pressure, $$V \propto T$$, so:

$$P \Delta V = n R \Delta T$$

Thus, the work done at constant pressure can be written as:

$$W = n R \Delta T$$

The heat supplied at constant pressure $$Q_p$$ is given by:

$$Q_p = n C_p \Delta T$$

Since the work done, $$W$$, at constant pressure is equivalent to $$n R \Delta T$$, it can also be related to $$Q_p$$ using the specific heat at constant pressure:

$$Q_p = W \cdot \frac{C_p}{R}$$

The specific heat at constant volume, $$Q_v$$, is given by:

$$Q_v = n C_v \Delta T$$

Since $$C_p / C_v = \gamma$$:

$$C_p = \gamma C_v$$

Therefore, we can substitute to get:

$$C_v = \frac{C_p}{\gamma}$$

Substitute $$C_v$$ into the expression for the heat supplied at constant volume:

$$Q_v = n \left(\frac{C_p}{\gamma}\right) \Delta T$$

Since $$n R \Delta T = W$$, and using the relationship between $$C_p$$ and $$R$$ for a monoatomic gas:

$$C_p = \frac{\gamma R}{\gamma - 1}$$

For a monoatomic gas, $$\gamma = \frac{5}{3}$$:

$$C_p = \frac{\frac{5}{3} R}{\frac{5}{3} - 1} = \frac{\frac{5}{3} R}{\frac{2}{3}} = \frac{5}{2} R$$

Therefore, $$C_v$$ is:

$$C_v = \frac{C_p}{\gamma} = \frac{\frac{5}{2} R}{\frac{5}{3}} = \frac{3R}{2}$$

Thus:

$$Q_v = n C_v \Delta T = n \frac{3R}{2} \Delta T$$

Since $$W = n R \Delta T$$, we can write:

$$Q_v = \frac{3}{2} \cdot W$$

Hence, the heat supplied at constant volume for the same rise in temperature of the gas is:

Option D: $$\frac{3 W}{2}$$

Heat And Thermodynamics

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