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Thermodynamics question

2020 · Shift 1 · Q7
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Thermodynamics question

2020 · Shift 1 · Q7

JEE AdvancedChemistryThermodynamicsMultiple correct+4 / −2
In thermodynamics, the p-V work done is given by w=−∫dVpextw = - \int {dV{p_{ext}}}w=−∫dVpext​ For a system undergoing a particular process, the work done is w=−∫dV(RTV−b−aV2)w = - \int {dV\left( {{{RT} \over {V - b}} - {a \over {{V^2}}}} \right)}w=−∫dV(V−bRT​−V2a​) This equation is applicable to a
  1. A
    system that satisfies the van der Walls' equation of state
  2. B
    process that is reversible and isothermal
  3. C
    process that is reversible and adiabatic
  4. D
    process that is irreversible and at constant pressure
View written solutionFree

Correct answer: A, B, C

  1. The general expression for pressure-volume work is

w=−∫pext dVw=-\int p_{\text{ext}}\,dVw=−∫pext​dV

This is valid for any mechanical process, where pextp_{\text{ext}}pext​ is the external pressure.

  1. In the given question, the work is written as

w=−∫dV(RTV−b−aV2)w=-\int dV\left(\frac{RT}{V-b}-\frac{a}{V^2}\right)w=−∫dV(V−bRT​−V2a​)

So here,

pext=RTV−b−aV2p_{\text{ext}}=\frac{RT}{V-b}-\frac{a}{V^2}pext​=V−bRT​−V2a​

  1. Recognize this pressure expression.

The van der Waals equation for 111 mole of a real gas is

(p+aV2)(V−b)=RT\left(p+\frac{a}{V^2}\right)(V-b)=RT(p+V2a​)(V−b)=RT

Rearranging,

p=RTV−b−aV2p=\frac{RT}{V-b}-\frac{a}{V^2}p=V−bRT​−V2a​

Thus the integrand is the van der Waals pressure of the gas.

So this equation is certainly applicable to a system obeying the van der Waals equation of state.

Hence, A is correct.

  1. Now check when we can replace pextp_{\text{ext}}pext​ by the system pressure ppp.

The work expression is always

w=−∫pext dVw=-\int p_{\text{ext}}\,dVw=−∫pext​dV

If the process is reversible, then at every stage

pext=psystemp_{\text{ext}}=p_{\text{system}}pext​=psystem​

Therefore,

w=−∫p dVw=-\int p\,dVw=−∫pdV

and if the system is a van der Waals gas,

p=RTV−b−aV2p=\frac{RT}{V-b}-\frac{a}{V^2}p=V−bRT​−V2a​

So the given expression applies to a reversible process for a van der Waals gas.

  1. Check option B: reversible and isothermal.

For an isothermal reversible process of a van der Waals gas, TTT is constant, so

w=−∫(RTV−b−aV2)dVw=-\int \left(\frac{RT}{V-b}-\frac{a}{V^2}\right)dVw=−∫(V−bRT​−V2a​)dV

is directly valid.

Hence, B is correct.

  1. Check option C: reversible and adiabatic.

For a reversible adiabatic process, we still have

w=−∫p dVw=-\int p\,dVw=−∫pdV

and for a van der Waals gas,

p=RTV−b−aV2p=\frac{RT}{V-b}-\frac{a}{V^2}p=V−bRT​−V2a​

So formally the same substitution is valid. In this case, however, TTT is not constant; it varies with VVV. Still, the integrand remains

RTV−b−aV2\frac{RT}{V-b}-\frac{a}{V^2}V−bRT​−V2a​

with TTT understood as the instantaneous temperature along the path. Therefore the given form is applicable to a reversible adiabatic process as well.

Hence, C is correct.

  1. Check option D: irreversible and at constant pressure.

For an irreversible constant-pressure process,

w=−pext∫dV=−pextΔVw=-p_{\text{ext}}\int dV=-p_{\text{ext}}\Delta Vw=−pext​∫dV=−pext​ΔV

Here the external pressure is constant and is generally not equal to

RTV−b−aV2\frac{RT}{V-b}-\frac{a}{V^2}V−bRT​−V2a​

throughout the path. The given expression uses the system pressure from the van der Waals equation, which is appropriate for reversible processes, not for a general irreversible constant-pressure process.

Hence, D is incorrect.

  1. Final answer:

The applicable options are

A, B, C\boxed{A,\ B,\ C}A, B, C​

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