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JEE Advance - Chemistry (2018 - Paper 1 Offline - No. 13)

A reversible cyclic process for an ideal gas is shown below. Here, $$P, V,$$ and $$T$$ are pressure, volume and temperature, respectively. The thermodynamic parameters $$q,w, H$$ and $$U$$ are heat, work, enthalpy and internal energy, respectively.

JEE Advanced 2018 Paper 1 Offline Chemistry - Thermodynamics Question 30 English

The correct option(s) is (are)
$${q_{Ac}} = \Delta {U_{BC}}\,\,$$ and $${w_{AB}} = {P_2}\left( {{V_2} - {V_1}} \right)$$
$${w_{BC}} = {P_2}\left( {{V_2} - {V_1}} \right)\,\,$$ and $${q_{BC}} = \Delta {H_{AC}}$$
$$\Delta {H_{CA}} < \Delta {U_{CA}}\,\,$$ and $$\,\,{q_{AC}} = \Delta {U_{BC}}$$
$$\,\,{q_{BC}} = \Delta {H_{AC}}\,\,$$ and $$\Delta {H_{CA}} > \Delta {U_{CA}}$$

Explanation

In the given curve AC represents isochoric process as volume at both the points is same i.e., V1

Similarly, AB represents isothermal process (as both the points are at T1 temperature) and BC represents isobaric process as both the points are at p2 pressure.

Now, (i) for option (a)

qAC = $$\Delta$$UBC = nCv(T2 $$-$$ T1)

where, n = number of moles

Cv = specific heat capacity at constant volume

However, WAB $$\ne$$ p2(V2 $$-$$ V1) instead

$${W_{AB}} = nR{T_1}\ln \left( {{{{V_2}} \over {{V_1}}}} \right)$$

So, this option is incorrect.

(ii) For option (b)

qBC = $$\Delta$$HAC = nCp(T2 $$-$$ T1)

where, Cp = specific heat capacity at constant pressure

Likewise,

WBC = $$-$$p2(V1 $$-$$ V2)

Hence, this option is correct.

(iii) For option (c)

as nCp(T2 $$-$$ T1) < nCv(T2 $$-$$ T1)

so, $$\Delta$$HCA < $$\Delta$$UCA

and qAC = $$\Delta$$UBC

Hence, this option is also correct.

(iv) For option (d)

Although qBC = $$\Delta$$HAC

but $$\Delta$$HCA $$\le$$ $$\Delta$$UCA

Hence, this option is incorrect.

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