INPhO 2021, Теория — Задача 2

Indian Olympiad Qualifier in Physics (IOQP) 2020-2021 — Part II: Indian National Physics Olympiad (INPhO) · 7 февруари 2021 г. · 8 т.

Указания

  1. This booklet consists of 4 pages and total of 5 questions. Write roll number at the top wherever asked.

  2. Booklet to write the answers is provided separately. Instructions to write the answers are on the Answer Booklet.

  3. Marks will be awarded on the basis of what you write on both the Summary Answer Sheet and the Detailed Answer Sheets in the Answer Booklet. Simple short answers and plots may be directly entered in the Summary Answer Sheet. Marks may be deducted for absence of detailed work in questions involving longer calculations.

  4. Strike out any rough work that you do not want to be considered for evaluation. You may also use the space on the Question Paper for rough work – this will NOT be evaluated.

  5. Non-programmable scientific calculators are allowed. Mobile phones cannot be used as calculators.

  6. Last page of the question paper can be used for rough work.

  7. Please submit the Answer Booklet at the end of the examination. You may retain the Question Paper.

Table of Constants
QuantitySymbolValue
Speed of light in vacuumcc3.00×108 m⋅s−13.00 \times 10^{8}\ \mathrm{m{\cdot}s^{-1}}
Planck's constanthh6.63×10−34 J⋅s6.63 \times 10^{-34}\ \mathrm{J{\cdot}s}
Magnitude of electron chargeee1.60×10−19 C1.60 \times 10^{-19}\ \mathrm{C}
Rest mass of electronmem_e9.11×10−31 kg9.11 \times 10^{-31}\ \mathrm{kg}
Value of 1/4πϵ01/4\pi\epsilon_09.00×109 N⋅m2⋅C−29.00 \times 10^{9}\ \mathrm{N{\cdot}m^{2}{\cdot}C^{-2}}
Acceleration due to gravitygg9.81 m⋅s−29.81\ \mathrm{m{\cdot}s^{-2}}
Note on solutions

Please note that alternate/equivalent methods and different way of expressing final solutions may exist. A correct method will be suitably awarded.

Условие

Prof. Saha gave the following problem to four students.

In this problem work done by a system on its surroundings is taken as positive. A non-ideal gas follows the Van der Waals equation of state

(P+n2aV2)(V−nb)=nRT\left(P + \frac{n^2 a}{V^2}\right)(V - nb) = nRT

where PP, VV, and TT denote the pressure, volume, and temperature, respectively; nn is the number of moles; RR is the universal gas constant and a,ba, b are dimensional positive constants. This gas expands adiabatically from an initial temperature TiT_i and volume ViV_i to a final temperature TfT_f and volume VfV_f. The adiabatic process is described by an equation of the form f(P,V;n,a,b,α)=f(P, V; n, a, b, \alpha) = constant, where α\alpha is a dimensionless number which is greater than 1. It is given that α→γ\alpha \to \gamma in the ideal gas limit, where γ\gamma is the adiabatic exponent. What is the work (WW) done by the gas in the process?

The four students solved the problem independently and gave four different answers. Their answers were:

(a) W=nRα−1(Ti−Tf)+n2a(Vf−1−Vi−1)W = \dfrac{nR}{\alpha - 1}(T_i - T_f) + n^2 a (V_f^{-1} - V_i^{-1})

(b) W=nRα−1(Tf−Ti)+n2a(Vf−1−Vi−1)W = \dfrac{nR}{\alpha - 1}(T_f - T_i) + n^2 a (V_f^{-1} - V_i^{-1})

(c) W=nRα−1(Ti−Tf)+n2a(Vfα−1−Viα−1)W = \dfrac{nR}{\alpha - 1}(T_i - T_f) + n^2 a (V_f^{\alpha - 1} - V_i^{\alpha - 1})

(d) W=nRα−1(Ti−Tf)[1−(Vf−nbVi−nb)α−1]W = \dfrac{nR}{\alpha - 1}(T_i - T_f)\left[1 - \left(\dfrac{V_f - nb}{V_i - nb}\right)^{\alpha - 1}\right]

Now, Prof. Saha had actually provided the exact expression of f(P,V;n,a,b,α)f(P, V; n, a, b, \alpha) to the students, but could not remember it during evaluation. Still, he could determine that some or all of the four answers above must be incorrect, based on general physical arguments alone.

Consider each of the four answers and give at least one reason for each of them showing why it is wrong, or possibly correct. Note that you are not required to give a correct expression for WW or a detailed derivation for it in this question.

Отговори

Решение

Оригинал в Архива: IOQP2021-PartII-Questions-en.pdf · официални решения: IOQP2021-PartII-Solutions-20210217.pdf