JEE Advance - Chemistry (2020 - Paper 1 Offline - No. 7)
In thermodynamics, the p-V work done is given by
$$w = - \int {dV{p_{ext}}} $$
For a system undergoing a particular process, the work done is
$$w = - \int {dV\left( {{{RT} \over {V - b}} - {a \over {{V^2}}}} \right)} $$
This equation is applicable to a
$$w = - \int {dV{p_{ext}}} $$
For a system undergoing a particular process, the work done is
$$w = - \int {dV\left( {{{RT} \over {V - b}} - {a \over {{V^2}}}} \right)} $$
This equation is applicable to a
system that satisfies the van der Walls' equation of state
process that is reversible and isothermal
process that is reversible and adiabatic
process that is irreversible and at constant pressure
Explanation
Given, $$w = - \int {{p_{ext}}dV} $$
For 1 mole van der Walls' gas
$$p = \left( {{{RT} \over {V - b}} - {a \over {{V^2}}}} \right)$$
For reversible process, pext = pgas
$$w = - \int {\left( {{{RT} \over {V - b}} - {a \over {{V^2}}}} \right)dV} $$
But, it is not applicable for irreversible process which are carried out very fast. So, work done is calculated assuming final pressure remains constant throughout the process. Thus, statement (a), (b) and (c) correct while statement (d) is incorrect.
For 1 mole van der Walls' gas
$$p = \left( {{{RT} \over {V - b}} - {a \over {{V^2}}}} \right)$$
For reversible process, pext = pgas
$$w = - \int {\left( {{{RT} \over {V - b}} - {a \over {{V^2}}}} \right)dV} $$
But, it is not applicable for irreversible process which are carried out very fast. So, work done is calculated assuming final pressure remains constant throughout the process. Thus, statement (a), (b) and (c) correct while statement (d) is incorrect.
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