Question

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

the correct option is (are) :
$\sf(a) \: q_{AC \: = \Delta \: U_{BC}} \: and \: W_{AB} = P_2 (V_2-V_1)$
$\sf (b) \: W_{BC} = P_2 (V_2-V_{1}) \: and \: a_{BC} = \Delta \: H_{AC}$
$\sf(c) \: \Delta \: H _{CA} < \Delta \: U_{CA} \: and \: q_{AC} = \Delta \: U_{BC}$
$\sf(d) \: q_{BC} = \Delta \: H _{AC} \: and \: \Delta \: H_{CA} > \Delta \: U_{CA}$

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Answer 1

Given, P= pressure, V = volume and T = temperature (in the given order).  

The thermodynamic parameters q, w, H and U are heat, work, enthalpy and internal energy, respectively.

Hi!! Ok so let's start verifying the provided options one by one :)

As we can conclude from the figure,

AB = Isothermal, BC = Isobaric and AC = Isochoric

(a) Part 1-  $\sf{\bf{qAC=\triangle U_{AB}}}$

As we know △U=q+w where, w = work done

$\implies\sf{\triangle U_{AC}=q_{AC}+w_{AC}}$

Putting, w = 0 because w = -P△V = 0 (AC = Isochoric)

$\implies\sf{\triangle U_{AC}=q_{AC}---(i)}$

$\implies\sf{\triangle U=0=\triangle U_{AB}+\triangle U_{BC}+\triangle U_{CA}}$

Putting, $\sf{\triangle U_{AB} = 0}$ (AB = Isothermal)

$\implies\sf{\triangle U_{BC}=-\triangle U_{CA}}$

And also,

$\implies\sf{\triangle U_{BC}=\triangle U_{AC}---(ii)}$

From (i) and (ii),

$\implies\bf{q_{AC}=\triangle U_{BC}}$

Hence, the first part of option (a) is correct.

Part 2 - $\bf{W_{AB}=P_2(V_2-V_1)}$

To find the work in isothermal usually we apply,

$\sf{W_{AB}=-RT\:ln\:\frac{V_2}{V_1}}$

Hence, the second part of option (a) is incorrect.

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(b) Part 1 - $\bf{W_{BC}=P_2(V_2-V_1)}$

$\implies\sf{W_{BC}=-P(\triangle V)}$

$\implies\sf{W_{BC}=-P_2(V_1-V_2)}$

$\implies\bf{W_{BC}=P_2(V_2-V_1)}$

Hence, the first part of option (b) is correct.

Part 2 -  $\bf{a_{BC}=\triangle H_{AC}}$

$\implies\sf{a_{BC}=\triangle H_{BC}---(iii)}$

from (ii),

$\implies\sf{\triangle H_{BC}=\triangle H_{AC}---(iv)}$

from (iii) and (iv),

$\implies\bf{a_{BC}=\triangle H_{AC}}$

Hence, the second part of option (b) is correct.

(P.S. It should be q instead of a:))

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(c) Part 1 - $\bf{\triangle H_{CA}<\triangle U_{CA}}$

$\implies\sf{\triangle H_{CA}=\triangle U_{CA}+\triangle PV}$

$\implies\sf{\triangle H_{CA}=\triangle U_{CA}+V_1(P_1-P_2)<0}$

$\implies\sf{\triangle H_{CA}-\triangle U_{CA}<0}$

$\implies\bf{\triangle H_{CA}<\triangle U_{CA}}$

Hence, the first part of option (c) is correct.

Part 2 - Already proved. (correct)

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(d) is just the opposite of (c) and hence it is incorrect.

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∴Option B and C are the correct options.