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

Question

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(4)

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

$Δ Q = 6 R T_{0}$

Solution

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

1 $2 R T_{0}$

2 $3 R T_{0}$

3 $4 R T_{0}$

4 $6 R T_{0}$

Ans.
(4)

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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Solution

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

1 2RT0

2 3RT0

3 4RT0

4 6RT0

Ans. (4) Net heat absorbed by one mole of diatomic gas in going from A→B (isochoric process) and B→C (isobaric process) is ΔQ=CvΔT+CpΔT=52RT0+72RT0 ΔQ=6RT0