One gram mole of an ideal gas at `N.T.P` is first expanded isothermally to twice the origional volume. It is then compressed at constant volume, till its pressure is raised to the original value. Calculate the total amount of work done. Given `R= 8.3 mol e^(_1)K^(-1)`.

Answer

To solve the problem step by step, we will calculate the work done during the isothermal expansion and the work done during the constant volume compression.

### Step 1: Understand the Initial Conditions
1. We have 1 gram mole of an ideal gas at Normal Temperature and Pressure (NTP).
2. At NTP, the temperature (T) is 300 K, and the pressure (P) is 1 atm (which is approximately 101.3 kPa).

### Step 2: Isothermal Expansion
1. The gas is expanded isothermally to twice its original volume.
- Let the original volume be \( V_1 \).
- After expansion, the new volume \( V_2 = 2V_1 \).

2. The work done (W) during an isothermal expansion can be calculated using the formula:
\[ W = nRT \ln \left( \frac{V_2}{V_1} \right) \]
where:
- \( n = 1 \) mole (since we have 1 gram mole of gas),
- \( R = 8.3 \, \text{J/mol K} \),
- \( T = 300 \, \text{K} \),
- \( V_2 = 2V_1 \).

3. Substituting the values into the formula:
\[ W = 1 \cdot 8.3 \cdot 300 \cdot \ln(2) \]
- We know that \( \ln(2) \approx 0.693 \).

4. Now calculate:
\[ W = 8.3 \cdot 300 \cdot 0.693 \approx 1720.5 \, \text{J} \]

### Step 3: Constant Volume Compression
1. The gas is then compressed at constant volume until its pressure returns to the original value.
2. In a constant volume process, the work done (W) is zero because there is no change in volume:
\[ W = 0 \]

### Step 4: Total Work Done
1. The total work done during the entire process is the sum of the work done during the isothermal expansion and the work done during the constant volume compression:
\[ W_{\text{total}} = W_{\text{isothermal}} + W_{\text{isochoric}} = 1720.5 \, \text{J} + 0 \, \text{J} = 1720.5 \, \text{J} \]

### Final Answer
The total amount of work done is approximately:
\[ \boxed{1720.5 \, \text{J}} \]