An ideal mono atomic gas A is supplied heat so as to expand without changing its temperature. In another process, starting with the same state, it is supplied heat at constant pressure. In both the cases a graph of work done by the gas (W) is plotted versus heat added (Q) to the gas. The ratio of slope of the graphs obtained in first and second process is `eta_(1)`. The same ratio obtained for an ideal diatomic gas in `eta_(2)`. Find the ratio `(eta_(1))/(eta_(2))`.

Answer

To solve the problem, we need to analyze the two processes of heat addition to an ideal monoatomic gas and an ideal diatomic gas, and find the ratio of the slopes of the work done (W) versus heat added (Q) graphs for both gases.

### Step-by-Step Solution:

1. **Understanding the Processes:**
- **Process 1 (Isothermal Expansion):** An ideal monoatomic gas is supplied heat to expand without changing its temperature. Since the temperature is constant, the internal energy change (ΔU) is zero. According to the first law of thermodynamics:
\[ Q = W + \Delta U \implies Q = W \quad (\text{since } \Delta U = 0) \]
- **Process 2 (Isobaric Process):** The gas is supplied heat at constant pressure. For an ideal gas, the heat added at constant pressure is given by:
\[ Q = nC_p\Delta T \]
where \(C_p\) is the specific heat at constant pressure.

2. **Finding the Work Done in Each Process:**
- For the isothermal process, since \(Q = W\), the slope \(m_1\) of the graph (W vs Q) is:
\[ m_1 = 1 \]
- For the isobaric process, we know that:
\[ W = Q - \Delta U \]
For an ideal gas, \(\Delta U = nC_v\Delta T\) and for a monoatomic gas, \(C_v = \frac{3}{2}R\). Thus:
\[ \Delta U = nC_v\Delta T = n\left(\frac{3}{2}R\right)\Delta T \]
Therefore, the work done \(W\) can be expressed as:
\[ W = Q - nC_v\Delta T \]
Since \(Q = nC_p\Delta T\) and \(C_p = C_v + R\), we have:
\[ C_p = \frac{5}{2}R \quad \text{(for monoatomic gas)} \]
Thus:
\[ W = nC_p\Delta T - nC_v\Delta T = n\left(C_p - C_v\right)\Delta T = nR\Delta T \]
The slope \(m_2\) for this process is:
\[ m_2 = \frac{W}{Q} = \frac{nR\Delta T}{nC_p\Delta T} = \frac{R}{C_p} = \frac{R}{\frac{5}{2}R} = \frac{2}{5} \]

3. **Finding the Ratios for Diatomic Gas:**
- For a diatomic gas, the specific heats are:
\[ C_v = \frac{5}{2}R \quad \text{and} \quad C_p = \frac{7}{2}R \]
- For the isothermal process, the slope remains the same:
\[ m_1 = 1 \]
- For the isobaric process:
\[ W = Q - \Delta U \]
where \(\Delta U = nC_v\Delta T = n\left(\frac{5}{2}R\right)\Delta T\). Thus:
\[ W = nC_p\Delta T - nC_v\Delta T = n\left(C_p - C_v\right)\Delta T = n\left(\frac{7}{2}R - \frac{5}{2}R\right)\Delta T = nR\Delta T \]
The slope \(m_2\) for the diatomic gas is:
\[ m_2 = \frac{W}{Q} = \frac{nR\Delta T}{nC_p\Delta T} = \frac{R}{C_p} = \frac{R}{\frac{7}{2}R} = \frac{2}{7} \]

4. **Finding the Ratio of Slopes:**
- The ratio of the slopes for the monoatomic gas to the diatomic gas is:
\[ \eta_1 = \frac{m_1}{m_2} = \frac{1}{\frac{2}{5}} = \frac{5}{2} \]
\[ \eta_2 = \frac{m_1}{m_2} = \frac{1}{\frac{2}{7}} = \frac{7}{2} \]
- Finally, the ratio \(\frac{\eta_1}{\eta_2}\) is:
\[ \frac{\eta_1}{\eta_2} = \frac{\frac{5}{2}}{\frac{7}{2}} = \frac{5}{7} \]

### Final Answer:
\[ \frac{\eta_1}{\eta_2} = \frac{5}{7} \]