One mole of ideal diatomic gas is taken from state $A \to B \to C$ as shown in P-T diagram. The total heat absorbed by the gas in the complete process is

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$2 R T_{0}$

$6 R T_{0}$

$4 R T_{0}$

$3 R T_{0}$

answer is D.

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Detailed Solution

Net heat absorbed by one mole of diatomic gas in going from
A $\to$ B (isochoric process) and B $\to$ C (isobaric process)
is
$Δ Q = C_{V} Δ T + C_{P} Δ T = \frac{5}{2} R T_{0} + \frac{7}{2} R T_{0}$
$Δ Q = 6 R T_{0}$

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