Question

Pleas state whether the following statement is TRUE or FALSE and explain your answer (using words or \equation)

1-The change in internal energy for going around the full cycle (Starting at point 1 and ending up back at point 1 again) will be zero.

2- Assuming that the a and b coefficients are approximately the same as those for the van der Waals equation a pressure calculated by the Berthelot would be less than calculated using the van der Waals equation.

Answer 1

STATEMENT 1 : TRUE

STATEMENT 2 : FALSE

Answer 2

Answers:

1. Change in Internal Energy for a Full Cycle

Statement:
"The change in internal energy for going around the full cycle (starting at point 1 and ending back at point 1) will be zero."

Answer: TRUE

Explanation:
Internal energy ($U$) is a state function, meaning it depends only on the current state of the system (e.g., pressure, volume, temperature), not the path taken. For a cyclic process (returning to the initial state), the net change in internal energy is zero:

$$\mathrm{\Delta }{U}_{\text{cycle}}=U_{\text{final}}-U_{\text{initial}}=0$$

This holds true regardless of the process (e.g., isothermal, adiabatic) during the cycle.

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2. Pressure Comparison: Berthelot vs. van der Waals Equation

Statement:
"Assuming the $a$ and $b$ coefficients are approximately the same for both equations, the pressure calculated by the Berthelot equation would be less than that calculated by the van der Waals equation."

Answer: FALSE

Explanation:
The Berthelot equation modifies the van der Waals equation by adding a temperature-dependent term to the attractive forces:

$$P_{\text{Berthelot}}=\frac{RT}{V_m-b}-\frac{a}{T{V}_m^2}$$

Compared to the van der Waals equation:

$$P_{\text{vdW}}=\frac{RT}{V_m-b}-\frac{a}{V_m^2}$$

The Berthelot term $\frac{a}{T{V}_m^2}$ is smaller than the van der Waals term $\frac{a}{V_m^2}$ (since $T>0$), resulting in a higher pressure for Berthelot at the same $V_m$.

Conclusion:
If $a$ and $b$ are similar, $P_{\text{Berthelot}}>P_{\text{vdW}}$. Thus, the statement is false.

Answer 3

Let's analyze each statement:

Statement 1:

"The change in internal energy for going around the full cycle (starting at point 1 and ending up back at point 1 again) will be zero."

Answer: TRUE

Explanation:

• Internal energy (U) is a state function. This means its value depends only on the current state of the system and not on the path taken to reach that state.

• In a cyclic process, the system returns to its initial state. Therefore, the initial and final internal energies are the same.

• The change in internal energy (ΔU) is calculated as:

  ΔU = U_final - U_initial

• Since U_final = U_initial in a cyclic process, we have:

  ΔU = U_initial - U_initial = 0

Statement 2:

"Assuming that the a and b coefficients are approximately the same as those for the van der Waals equation a pressure calculated by the Berthelot would be less than calculated using the van der Waals equation."

Answer: FALSE

Explanation:

• Van der Waals Equation:

  (P+V2an2)(V−nb)=nRT

  where:

• P = pressure

• V = volume

• n = number of moles

• R = ideal gas constant

• T = temperature

• a and b are constants specific to the gas

• Berthelot Equation:

  P=V−nbnRT−TV2an2

• Comparison:

  Let's rearrange both equations to solve for P:

  Van der Waals:

  P=V−nbnRT−V2an2

  Berthelot:

  P=V−nbnRT−TV2an2

• The only difference between the two equations is the temperature term in the attractive force correction term (the term with 'a'). In the Berthelot equation, it's divided by T, while in the van der Waals equation, it's not.

• Since temperature (T) is always positive, dividing by T will make the attractive force correction term smaller in the Berthelot equation compared to the van der Waals equation.

• Therefore, the Berthelot equation will result in a higher pressure than the van der Waals equation, not lower.

In summary:

1. The change in internal energy for a cyclic process is indeed zero.

2. The Berthelot equation, under the given conditions, predicts a higher pressure than the van der Waals equation.