One mole of a monatomic ideal gas has an initial pressure, volume, and temperature of `P_0 , V_0`, and 438 K, respectively. It undergoes an isothermal expansion that triples the volume of the gas. Then, the gas undergoes an isobaric compression back to its original volume. Finally, the gas undergoes an isochoric increase in pressure, so that the final pressure, volume, and temperature are `p_0 V_0 ` and 438 K, respectively. Find the total heat for this three-step process, and state whether it is absorbed by or given off by the gas.

Options

A. `3640 J` absorbed
B. `-1820 ` J given off
C. 1570 J , absorbed
D. `-1110 J` , given off

Answer

To solve the problem step by step, we will analyze each process the gas undergoes and calculate the total heat exchanged during the entire cycle.

### Step 1: Identify the processes and their characteristics
1. **Isothermal Expansion (Process 1-2)**: The gas expands at a constant temperature (438 K), and its volume triples from \( V_0 \) to \( 3V_0 \).
2. **Isobaric Compression (Process 2-3)**: The gas is compressed at constant pressure back to its original volume \( V_0 \).
3. **Isochoric Process (Process 3-1)**: The gas pressure increases at constant volume back to its original state.

### Step 2: Calculate the work done in each process

#### Process 1-2 (Isothermal Expansion)
The work done during an isothermal process for an ideal gas is given by:
\[ W_{12} = nRT \ln\left(\frac{V_f}{V_i}\right) \]
Where:
- \( n = 1 \) mole (given)
- \( R = 8.314 \, \text{J/(mol K)} \)
- \( T = 438 \, \text{K} \)
- \( V_f = 3V_0 \)
- \( V_i = V_0 \)

Substituting the values:
\[ W_{12} = 1 \times 8.314 \times 438 \ln\left(\frac{3V_0}{V_0}\right) = 8.314 \times 438 \ln(3) \]

#### Process 2-3 (Isobaric Compression)
The work done during an isobaric process is given by:
\[ W_{23} = P \Delta V \]
Since the pressure remains constant, we first need to find the pressure at state 2. From the ideal gas law:
\[ P_2 = \frac{nRT}{V_2} = \frac{1 \times 8.314 \times 438}{3V_0} = \frac{8.314 \times 438}{3V_0} \]
The change in volume is:
\[ \Delta V = V_f - V_i = V_0 - 3V_0 = -2V_0 \]
Thus,
\[ W_{23} = P_2 \Delta V = \left(\frac{8.314 \times 438}{3V_0}\right)(-2V_0) = -\frac{2 \times 8.314 \times 438}{3} \]

#### Process 3-1 (Isochoric Process)
In an isochoric process, the work done is zero:
\[ W_{31} = 0 \]

### Step 3: Calculate total work done in the cycle
The total work done \( W_{total} \) is:
\[ W_{total} = W_{12} + W_{23} + W_{31} \]
Substituting the values:
\[ W_{total} = 8.314 \times 438 \ln(3) - \frac{2 \times 8.314 \times 438}{3} + 0 \]

### Step 4: Apply the First Law of Thermodynamics
According to the First Law of Thermodynamics:
\[ \Delta U = Q - W \]
For a complete cycle, the change in internal energy \( \Delta U = 0 \). Therefore:
\[ 0 = Q_{total} - W_{total} \implies Q_{total} = W_{total} \]

### Step 5: Determine the sign of heat transfer
If \( Q_{total} > 0 \), heat is absorbed by the gas; if \( Q_{total} < 0 \), heat is given off by the gas.

### Final Calculation
Now, we can compute the numerical values for \( Q_{total} \) based on the calculations above.

1. Calculate \( W_{12} \):
\[ W_{12} = 8.314 \times 438 \ln(3) \approx 8.314 \times 438 \times 1.0986 \approx 3,236.7 \, \text{J} \]

2. Calculate \( W_{23} \):
\[ W_{23} = -\frac{2 \times 8.314 \times 438}{3} \approx -2,426.2 \, \text{J} \]

3. Total Work Done:
\[ W_{total} = 3,236.7 - 2,426.2 = 810.5 \, \text{J} \]

Thus, the total heat \( Q_{total} \) is approximately:
\[ Q_{total} \approx 810.5 \, \text{J} \]
Since this value is positive, the gas absorbs heat.

### Summary
- Total heat absorbed by the gas: \( Q_{total} \approx 810.5 \, \text{J} \)
- The gas absorbs heat during the process.