A gas `(C_(v.m)=(5)/(2)R)` behaving ideally was allowed to expand reversibly and adiabatically from 1 liter to
32 liter . It's initial at temperature was `327^(@)` C . The molar enthalpy changes `("in "J//"mole")` for the process is

Explanation

To solve the problem, we need to find the molar enthalpy change (ΔHm) for a gas that expands reversibly and adiabatically from 1 liter to 32 liters, starting at a temperature of 327°C. The specific heat at constant volume (Cv) is given as (5/2)R.

### Step-by-step Solution:

1. **Convert Initial Temperature to Kelvin:**
\[ T_1 = 327°C = 327 + 273 = 600 \, K \]

**Hint:** Remember to convert Celsius to Kelvin by adding 273.

2. **Calculate Cp (Specific Heat at Constant Pressure):**
\[ C_p = C_v + R = \frac{5}{2}R + R = \frac{5}{2}R + \frac{2}{2}R = \frac{7}{2}R \]

**Hint:** Use the relationship \( C_p = C_v + R \) to find Cp.

3. **Determine the Ratio of Volumes:**
\[ V_1 = 1 \, L, \quad V_2 = 32 \, L \]
\[ \frac{V_1}{V_2} = \frac{1}{32} \]

**Hint:** Identify the initial and final volumes to find their ratio.

4. **Calculate the Value of γ (Gamma):**
\[ \gamma = \frac{C_p}{C_v} = \frac{\frac{7}{2}R}{\frac{5}{2}R} = \frac{7}{5} \]

**Hint:** Use the formula \( \gamma = \frac{C_p}{C_v} \) to find gamma.

5. **Use the Adiabatic Relation to Find T2:**
\[ \frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{\gamma - 1} = \left(\frac{1}{32}\right)^{\frac{7}{5} - 1} = \left(\frac{1}{32}\right)^{\frac{2}{5}} \]
\[ T_2 = T_1 \left(\frac{1}{32}\right)^{\frac{2}{5}} = 600 \left(\frac{1}{32}\right)^{\frac{2}{5}} \]

**Hint:** Use the adiabatic process relation \( T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{\gamma - 1} \).

6. **Calculate \( T_2 \):**
\[ T_2 = 600 \times \left(\frac{1}{32}\right)^{0.4} = 600 \times \frac{1}{4} = 150 \, K \]

**Hint:** Simplify the calculation step by step, ensuring to evaluate the exponent correctly.

7. **Calculate the Change in Temperature (ΔT):**
\[ \Delta T = T_2 - T_1 = 150 \, K - 600 \, K = -450 \, K \]

**Hint:** Remember that ΔT is the final temperature minus the initial temperature.

8. **Calculate the Molar Enthalpy Change (ΔHm):**
\[ \Delta H_m = C_p \Delta T = \frac{7}{2}R \times (-450) \]
\[ \Delta H_m = -1575R \, J/mole \]

**Hint:** Substitute the values of Cp and ΔT into the enthalpy change formula.

### Final Answer:
\[ \Delta H_m = -1575R \, J/mole \]