Pressure of 10 moles of an ideal gas is changed from 2 atm to 1 at against constant external pressure without change in temperature. If surrounding temperature (300K) and pressure (1 atm) always remains constant then calculate total entropy change `(DeltaS_("system")+DeltaS_("surrounding"))` for given process.
[Given : ? n2 = 0.70 and R = 8.0 J/mol/K]

Options

A. 56 J/K
B. 14 J/K
C. 16 J/K
D. None of these

Answer

To solve the problem of calculating the total entropy change (ΔS_system + ΔS_surroundings) for the given process, we will follow these steps:

### Step 1: Calculate ΔS_system
The formula for the change in entropy of the system (ΔS_system) for an ideal gas undergoing an isothermal process is given by:

\[ \Delta S_{system} = nR \ln\left(\frac{P_1}{P_2}\right) \]

Where:
- \( n = 10 \) moles (given)
- \( R = 8.0 \, \text{J/mol/K} \) (given)
- \( P_1 = 2 \, \text{atm} \) (initial pressure)
- \( P_2 = 1 \, \text{atm} \) (final pressure)

Substituting the values into the formula:

\[ \Delta S_{system} = 10 \times 8.0 \times \ln\left(\frac{2}{1}\right) \]

Calculating the natural logarithm:

\[ \ln(2) \approx 0.693 \]

Now substituting this value:

\[ \Delta S_{system} = 10 \times 8.0 \times 0.693 \approx 55.44 \, \text{J/K} \]

### Step 2: Calculate ΔS_surroundings
The change in entropy of the surroundings (ΔS_surroundings) can be calculated using the formula:

\[ \Delta S_{surroundings} = -\frac{q_{rev}}{T} \]

Where \( q_{rev} \) is the heat exchanged in a reversible process. For an isothermal process, the heat exchanged can be calculated as:

\[ q_{rev} = nRT \ln\left(\frac{P_2}{P_1}\right) \]

Substituting the values:

\[ q_{rev} = 10 \times 8.0 \times 300 \times \ln\left(\frac{1}{2}\right) \]

Calculating the natural logarithm:

\[ \ln\left(\frac{1}{2}\right) = -\ln(2) \approx -0.693 \]

Now substituting this value:

\[ q_{rev} = 10 \times 8.0 \times 300 \times (-0.693) \approx -16632 \, \text{J} \]

Now, substituting \( q_{rev} \) into the ΔS_surroundings formula:

\[ \Delta S_{surroundings} = -\frac{-16632}{300} \approx 55.44 \, \text{J/K} \]

### Step 3: Calculate Total Entropy Change
Now we can find the total entropy change by adding ΔS_system and ΔS_surroundings:

\[ \Delta S_{total} = \Delta S_{system} + \Delta S_{surroundings} \]

Substituting the values:

\[ \Delta S_{total} = 55.44 + 55.44 = 110.88 \, \text{J/K} \]

### Final Result
Thus, the total entropy change for the process is:

\[ \Delta S_{total} \approx 110.88 \, \text{J/K} \]