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Physics-

General

Easy

Question

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

$0, R T_{2} \ln (\frac{V_{1}}{V_{2}}), R (T_{1} - T_{2})$
$R (T_{1} - T_{2}), 0, R T_{1} \ln \frac{V_{1}}{V_{2}}$
$0, R T_{2} \ln (\frac{V_{2}}{V_{1}}), R (T_{1} - T_{2})$
$0, R T_{2} \ln (\frac{V_{2}}{V_{1}}), R (T_{2} - T_{1})$

The correct answer is: $0 comma R T subscript 2 end subscript ln invisible function application open parentheses fraction numerator V subscript 2 end subscript over denominator V subscript 1 end subscript end fraction close parentheses comma R left parenthesis T subscript 1 end subscript minus T subscript 2 end subscript right parenthesis$

Process AB is isochoric, \ $W subscript A B end subscript equals P capital delta V equals 0$
Process BC is isothermal\ $W subscript B C end subscript equals R T subscript 2 end subscript. ln invisible function application open parentheses fraction numerator V subscript 2 end subscript over denominator V subscript 1 end subscript end fraction close parentheses$
Process CA is isobaric
\ $W subscript C A end subscript equals negative P capital delta V equals negative R capital delta T equals negative R left parenthesis T subscript 1 end subscript minus T subscript 2 end subscript right parenthesis equals R left parenthesis T subscript 2 end subscript minus T subscript 1 end subscript right parenthesis$
(Negative sign is taken because of compression)

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