A cyclinder containing one gram mole of a gas was put on boiling water bath and compressed adibatically till its temperature rose by `70^(@)C`. Calculate the work done and heat developed in the gas, `gamma= 1.5, R= 2cal. mol e^(-1)K^(-1)`

Answer

To solve the problem, we need to calculate the work done and heat developed in the gas during the adiabatic compression. We are given the following data:

- Number of moles (n) = 1 mole
- Change in temperature (ΔT) = 70°C = 70 K (since the change in Celsius is equal to the change in Kelvin)
- Heat capacity ratio (γ) = 1.5
- Gas constant (R) = 2 cal/(mol·K)

### Step-by-Step Solution:

1. **Convert the change in temperature to Kelvin**:
\[ \Delta T = 70 \text{°C} = 70 \text{ K} \]

2. **Calculate the initial temperature (T1)**:
Since the gas is put in a boiling water bath, we can assume the initial temperature (T1) is the boiling point of water:
\[ T_1 = 100 \text{°C} = 373 \text{ K} \]

3. **Calculate the final temperature (T2)**:
The final temperature after the adiabatic compression is:
\[ T_2 = T_1 + \Delta T = 373 \text{ K} + 70 \text{ K} = 443 \text{ K} \]

4. **Calculate the work done (W) during adiabatic compression**:
The work done in an adiabatic process can be calculated using the formula:
\[ W = nR(T_2 - T_1) \frac{1}{\gamma - 1} \]
Substituting the values:
\[ W = 1 \text{ mol} \times 2 \text{ cal/(mol·K)} \times (443 \text{ K} - 373 \text{ K}) \frac{1}{1.5 - 1} \]
\[ W = 1 \times 2 \times 70 \times \frac{1}{0.5} = 2 \times 70 \times 2 = 280 \text{ cal} \]

5. **Calculate the change in internal energy (ΔU)**:
For an ideal gas, the change in internal energy during an adiabatic process is given by:
\[ \Delta U = nC_v \Delta T \]
Where \(C_v = \frac{R}{\gamma - 1}\):
\[ C_v = \frac{2 \text{ cal/(mol·K)}}{1.5 - 1} = \frac{2}{0.5} = 4 \text{ cal/(mol·K)} \]
Now, substituting the values:
\[ \Delta U = 1 \text{ mol} \times 4 \text{ cal/(mol·K)} \times 70 \text{ K} = 280 \text{ cal} \]

6. **Calculate the heat developed (Q)**:
In an adiabatic process, the heat exchanged (Q) is zero:
\[ Q = 0 \text{ cal} \]

### Final Answers:
- Work done (W) = 280 cal
- Heat developed (Q) = 0 cal