A cyclic process for 1, mole of an ideal gas is shown in figure in V-T, diagram. work done in AB, BC

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11.Thermodynamics

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A cyclic process for $1\, mole$ of an ideal gas is shown in figure in the $V-T,$ diagram. The work done in $AB, BC$ and $CA$ respectively

A

$0,\,R{T_2}\ln \left( {\frac{{{V_1}}}{{{V_2}}}} \right)\,,\,R\,({T_1} - {T_2})$

B

$R({T_1} - {T_2}),\,0,\,R{T_1}\ln \frac{{{V_1}}}{{{V_2}}}$

C

$0,\,R{T_2}\ln \left( {\frac{{{V_2}}}{{{V_1}}}} \right)\,,\,R\,({T_1} - {T_2})$

D

$0,\,R{T_2}\ln \left( {\frac{{{V_2}}}{{{V_1}}}} \right)\,,\,R\,({T_2} - {T_1})$

Solution

(c) Process $AB$ is isochoric, $\therefore$ ${W_{AB}} = P\,\Delta V = 0$

Process $BC$ is isothermal $\therefore$ ${W_{BC}} = R{T_2}.\ln \left( {\frac{{{V_2}}}{{{V_1}}}} \right)$

Process $CA$ is isobaric

$\therefore {W_{CA}} = – \,P\Delta V$$ = – \,R\Delta T$$ = – \,R({T_1} – {T_2})$$ = R({T_2} – {T_1})$

(Negative sign is taken because of compression)

Standard 11

Physics

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