A) Work in the reversible isothermal expansion of an ideal gas is less than for a van der Waals gas.
B) The Joule-Thomson coefficient for an ideal gas is zero.
C) Criterion of spontaneity reaction is: (i) \[{{(dS)}_{U,V}}\ge 0\] (ii) \[{{(dG)}_{T,P}}\le 0\] (iii) \[{{(dU)}_{S.V}}\le 0\] (iv) \[{{(dh)}_{S.P}}\le 0\] Where in equality (< or >) refers to an irreversible process (Spontaneous) while equality (=) refers to a reversible process at equilibrium.
D) Endergonic reaction is a reaction for which \[\Delta G=+ve\] and for exergonic reaction \[\Delta G=-ve\]
Correct Answer: A
Solution :
[a] Statement (1) is INCORRECT:- CORRECT: \[{{w}_{ideal}}>{{w}_{vdw}}\] as shown below: \[-{{w}_{ideal}}=nRT\] In \[\left( \frac{{{V}_{2}}}{{{V}_{1}}} \right)\] \[-{{w}_{vdw}}=nRT\] In \[\left( \frac{{{V}_{2}}}{{{V}_{1}}} \right)+a{{n}^{2}}\left( \frac{1}{{{V}_{2}}}-\frac{1}{{{V}_{1}}} \right)\] \[\therefore \,\,\,\,{{w}_{ideal}}-{{w}_{vdw}}=-a{{n}^{2}}\left( \frac{1}{{{V}_{2}}}-\frac{1}{{{V}_{1}}} \right)\] \[=-a{{n}^{2}}\left( \frac{{{V}_{1}}-{{V}_{2}}}{{{V}_{1}}{{V}_{2}}} \right)=a{{n}^{2}}\left( \frac{{{V}_{2}}-{{V}_{1}}}{{{V}_{1}}{{V}_{2}}} \right)\] ?..(i) Since for the expansion of a gas, \[{{V}_{2}}>{{V}_{1}},\] it is evident from equation (i), that numerically the work in the reversible isothermal expansion of an ideal gas is greater than for a van der Waals gas. An ideal gas has no intermolecular forces whereas a real (van der Waals) gas has considerable intermolecular forces. Thus in an ideal gas, the heat supplied is fully utilized in doing the work of expansion whereas in a real gas, a part of the heat supplied is used in overcoming the intermolecular forces of attraction between the molecules and the balance amount of heat is utilized in doing work of expansion consequently, numerically, \[{{w}_{ideal}}>{{w}_{vdw}}\] Statement (2) is CORRECT The change of temperature produced when a gas is made to expand adiabatically from a region of high pressure to a region of extremely low pressure is known as Joule- Thomson effect. (J.T.E). The cooling effect is due to decrease in kinetic energy of the gaseous molecules since a part of this energy is used up in overcoming the van der Waals foces of attraction existing between the molecules during expansion. J.T.E is very small when a gas approaches ideal behaviour thus J.T.E is zero in an ideal gas. Joule- Thomson coefficient \[({{\mu }_{J.T.}})\]in an ideal gas, Since \[H=U+PV\] \[\therefore \,\,\,\frac{\partial T}{\partial P}={{\mu }_{J.T.}}=\frac{1}{{{C}_{P}}}\left( \frac{\partial (U+PV)}{\partial P} \right)\] \[=-\frac{1}{{{C}_{P}}}\left[ {{\left( \frac{\partial U}{\partial V} \right)}_{T}}+{{\left( \frac{\partial (PV)}{\partial P} \right)}_{T}} \right]\] \[=-\frac{1}{{{C}_{P}}}\left[ {{\left( \frac{\partial U}{\partial V}\times \frac{\partial V}{\partial P} \right)}_{T}}+{{\left( \frac{\partial (PV)}{\partial P} \right)}_{T}} \right]\] \[=-\frac{1}{{{C}_{P}}}\left[ {{\left( \frac{\partial U}{\partial V} \right)}_{T}}{{\left( \frac{\partial V}{\partial P} \right)}_{T}}+{{\left( \frac{\partial (PV)}{\partial P} \right)}_{T}} \right]\] Since for an ideal gas \[{{\left( \frac{\partial U}{\partial V} \right)}_{T}}=0\] \[\therefore \,\,\,\,\,\,{{\left( \frac{\partial U}{\partial V} \right)}_{T}}{{\left( \frac{\partial V}{\partial V} \right)}_{T}}=0\] Also, for an ideal gas, PV is constant at constant T, \[\therefore \,\,\,\,\,\,\,\,\,\,{{\left( \frac{\partial (PV)}{\partial P} \right)}_{T}}=0\] \[{{\mu }_{J.T.}}=0\] Statement (3) and (4) are CORRECT
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