An ideal monoatomic gas is carried along the cycle ABCDA as shown in the figure. The total heat absorbed during this process is

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$2.5 P_{0} V_{0}$

$1.5 P_{0} V_{0}$

$10.5 P_{0} V_{0}$

$7.5 P_{0} V_{0}$

answer is A.

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Detailed Solution

Heat absorbed means heat is extracted from source, i.e. Q must be positive. The occurs along path
$A \to B \to C$
$∴$ Heat added = $Δ Q_{A B C} = Δ U_{A B C} + Δ w_{A B C}$
$= C_{v} (T_{C} - T_{A}) + 2 p_{0} (2 V_{0} - V_{0}) = n \frac{3}{2} R (T_{C} - T_{A}) + 3 p_{0} V_{0} = \frac{3}{2} (n R T_{C} - n R T_{A}) + 3 p_{0} - V_{0} = \frac{3}{2} (p_{C} V_{C} - p_{A} V_{A}) + 3 p_{0} V_{0} = \frac{3}{2} (3 p_{0} 2 V_{0} - p_{0} V_{0}) + 3 p_{0} V_{0} \frac{3}{2} \times {}^{3}P_{0} V_{0} + {}^{3}P_{0} V_{0} = \frac{21}{2} p_{0} V_{0} = 10.5 p_{0} V_{0}$