An ideal gas undergoes an isobaric process. If its heat capacity is `C`, at constant volume and number of mole `n,` then the ratio of work done by gas to heat given to gas when temperature of gas changes by `DeltaT` is :

Options

A. `((nR)/(C_(v)+R))`
B. `((R)/(C_(v)+R))`
C. `((nR)/(C_(v)-R))`
D. `((R)/(C_(v)-R))`

Answer

To solve the problem, we need to find the ratio of work done by an ideal gas to the heat given to it during an isobaric process when the temperature changes by \(\Delta T\).

### Step-by-Step Solution:

1. **Understand the Isobaric Process:**
In an isobaric process, the pressure \(P\) remains constant. For an ideal gas, the relationship between pressure, volume, and temperature is given by the ideal gas law:
\[ PV = nRT \]

2. **Calculate the Work Done by the Gas:**
The work done \(W\) by the gas during an isobaric process can be expressed as:
\[ W = P \Delta V \]
Using the ideal gas law, we can express \(\Delta V\) in terms of \(\Delta T\):
\[ \Delta V = \frac{nR \Delta T}{P} \]
Therefore, substituting this into the work equation gives:
\[ W = P \left(\frac{nR \Delta T}{P}\right) = nR \Delta T \]

3. **Calculate the Heat Given to the Gas:**
The heat \(Q\) given to the gas in an isobaric process can be calculated using:
\[ Q = \Delta U + W \]
where \(\Delta U\) is the change in internal energy. For an ideal gas, the change in internal energy is given by:
\[ \Delta U = nC_V \Delta T \]
Thus, substituting for \(Q\):
\[ Q = nC_V \Delta T + nR \Delta T = n(C_V + R) \Delta T \]

4. **Find the Ratio of Work Done to Heat Given:**
Now we can find the ratio of work done \(W\) to heat given \(Q\):
\[ \frac{W}{Q} = \frac{nR \Delta T}{n(C_V + R) \Delta T} \]
The \(\Delta T\) and \(n\) cancel out:
\[ \frac{W}{Q} = \frac{R}{C_V + R} \]

5. **Final Result:**
Therefore, the ratio of work done by the gas to the heat given to the gas is:
\[ \frac{W}{Q} = \frac{R}{C_V + R} \]