Sure, let's go through the question step-by-step.
1. Given Data:
- Moles of gas ([tex]\( n \)[/tex]): [tex]\( 5.0 \times 10^{-3} \)[/tex] moles
- Initial volume ([tex]\( V_i \)[/tex]): [tex]\( 0.0124 \)[/tex] dm[tex]\(^3\)[/tex], which is [tex]\( 0.0124 \times 10^{-3} \)[/tex] m[tex]\(^3\)[/tex]
- Final volume ([tex]\( V_f \)[/tex]): [tex]\( 2 \times V_i \)[/tex]
- Temperature ([tex]\( T \)[/tex]): 298 K
- Universal gas constant ([tex]\( R \)[/tex]): 8.314 J/(mol K)
- External pressure ([tex]\( P_\text{ext} \)[/tex]): 1.0 atm, which is 101325 Pa or 101325 N/m[tex]\(^2\)[/tex]
2. Final Volume:
[tex]\[
V_f = 2 \times V_i = 2 \times 0.0124 \text{ dm}^3 = 0.0248 \text{ dm}^3 = 0.0248 \times 10^{-3} \text{ m}^3
\][/tex]
3. Reversible Work:
For an isothermal reversible expansion:
[tex]\[
W_\text{rev} = -nRT \ln \left( \frac{V_f}{V_i} \right)
\][/tex]
Substituting the given values:
[tex]\[
W_\text{rev} = - (5.0 \times 10^{-3} \, \text{mol}) \times (8.314 \, \text{J/(mol K)}) \times (298 \, \text{K}) \times \ln \left( \frac{0.0248}{0.0124} \right)
\][/tex]
[tex]\[
\ln \left( \frac{0.0248}{0.0124} \right) = \ln (2) \approx 0.693
\][/tex]
[tex]\[
W_\text{rev} \approx - (5.0 \times 10^{-3}) \times 8.314 \times 298 \times 0.693 \approx -8.5866 \, \text{J}
\][/tex]
4. Heat Absorbed by System ([tex]\( q_\text{rev} \)[/tex]):
Under reversible conditions, for isothermal processes:
[tex]\[
q_\text{rev} = - W_\text{rev} = 8.5866 \, \text{J}
\][/tex]
5. Entropy Change of the System ([tex]\( \Delta S_\text{syst} \)[/tex]):
[tex]\[
\Delta S_\text{syst} = \frac{q_\text{rev}}{T} = \frac{8.5866 \, \text{J}}{298 \, \text{K}} \approx 0.0288 \, \text{J/K}
\][/tex]
6. Entropy Change of the Surroundings ([tex]\( \Delta S_\text{sur} \)[/tex]):
Given that [tex]\( \Delta S_\text{sur} = - \frac{q_\text{sys}}{T} \)[/tex],
[tex]\[
\Delta S_\text{sur} = - \frac{8.5866 \, \text{J}}{298 \, \text{K}} \approx -0.0288 \, \text{J/K}
\][/tex]
The negative sign indicates the heat lost by the surroundings, as it is heat gained by the system.
7. Total Entropy Change ([tex]\( \Delta S_\text{total} \)[/tex]):
[tex]\[
\Delta S_\text{total} = \Delta S_\text{syst} + \Delta S_\text{sur} = 0.0288 \, \text{J/K} + (-0.0288 \, \text{J/K}) = 0 \, \text{J/K}
\][/tex]
Conclusion:
The total entropy change ([tex]\( \Delta S_\text{total} \)[/tex]) is zero, consistent with the Second Law of Thermodynamics, which states that the entropy of an isolated system either increases or remains constant, but never decreases. In this case, the system and surroundings together are not isolated but are exchanging heat in a reversible manner, resulting in no net change in entropy.
