One mole of an ideal monoatomic gas is taken round cyclic process $A B C A$ as shown in figure. Calculate the net heat absorbed by the gas in the path $B C$ .

\frac{1}{3} P_{o} V_{o}

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\frac{P_{o} V_{o}}{2}

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P_{o} V_{o}

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2 P_{o} V_{o}

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Solution

The correct option is B $\frac{P_{o} V_{o}}{2}$
Work done by gas
W = area( $Δ$ ABC)
$= \frac{1}{2} (3 V_{o} - V_{o}) (3 P_{o} - P_{o}) = 2 P_{o} V_{o}$

Heat absorbed in path AB (isochoric)

$d θ_{A B} = C_{V} d T = C_{V} (T_{f i n a l} - T_{i n i t i a l})$

$= C_{V} (\frac{P_{f} V_{f}}{R} - \frac{P_{i} V_{i}}{R})$

$= \frac{C_{V}}{R} (P_{f} V_{f} - P_{i} V_{i})$

$= \frac{3}{2} (P_{f} V_{f} - P_{i} V_{i})$

$= \frac{3}{2} (3 P_{o} V_{o} - P_{o} V_{o}) = 3 P_{o} V_{o}$

Heat absorbed in path CA (isobaric)

$d θ_{C A} = C_{P} d T$
$= C_{P} (T_{f i n a l} - T_{i n i t i a l})$

$= C_{P} (\frac{P_{f} V_{f}}{R} - \frac{P_{i} V_{i}}{R})$

$= \frac{C_{P}}{R} (P_{f} V_{f} - P_{i} V_{i}) [C_{P} = \frac{5}{2} R \Rightarrow \frac{C_{P}}{R} = \frac{5}{2}] = \frac{5}{2} (P_{o} V_{o} - 3 P_{o} V_{o}) = - 5 P_{o} V_{o}$

(heat is released)
Total heat absorbed during the process

$d θ = d θ_{A B} + d θ_{B C} + d θ_{C A}$

$= 3 P_{o} V_{o} + d θ_{B C} + (- 5 P_{o} V_{o})$

$d θ = - 2 P_{o} V_{o} + d θ_{B C}$

Since change in internal energy is 0 i.e. $d U = 0$

$d θ = d W = 2 P_{o} V_{o}$

$2 P_{o} V_{o} = - 2 P_{o} V_{o} + d θ_{B C}$

$d θ_{B C} = 4 P_{o} V_{o}$

So heat absorbed by process BC is $\frac{P_{o} V_{o}}{2}$
Hence option B is correct.

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