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Standard X

Physics

Principle of Calorimetry

70 cal of hea...

Question

$70 c a l$ of heat is required to raise the temperature of a diatomic gas at constant pressure from $30 t o 35^{o} C$ . The amount of heat required (in cal) to raise the temperature of the same gas through the same range ( $30 t o 35^{o} C$ ) at constant volume is:

30

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60

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50

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70

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Solution

The correct option is C $50$
$\frac{C_{p}}{C_{v}} = γ = 1.4$
$\frac{{Δ Q}_{p}}{{Δ Q}_{v}} = \frac{n C_{p} Δ T}{n C_{v} Δ T} = \frac{C_{p}}{C_{v}} = 1.4$
$∴ {Δ Q}_{v} = \frac{{Δ Q}_{p}}{1.4} = 50 c a l$

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70 calories of heat are required to raise the temperature of 2 moles of an ideal gas at constant pressure from $30^{\circ} C t o 35^{\circ} C$ . The amount of heat required to raise the temperature of same gas through the same range ( $30^{\circ} C t o 35^{\circ} C$ ) at constant volume (R = 2 cal/mol/K)

70 cal of heat is required to raise the temperature of a diatomic gas at constant pressure from 30 to 35oC. The amount of heat required (in cal) to raise the temperature of the same gas through the same range (30 to 35oC) at constant volume is:

The correct option is C 50CpCv=γ=1.4ΔQpΔQv=nCpΔTnCvΔT=CpCv=1.4∴ΔQv=ΔQp1.4=50 cal

$70 c a l$ of heat is required to raise the temperature of a diatomic gas at constant pressure from $30 t o 35^{o} C$ . The amount of heat required (in cal) to raise the temperature of the same gas through the same range ( $30 t o 35^{o} C$ ) at constant volume is:

The correct option is C $50$
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$\frac{{Δ Q}_{p}}{{Δ Q}_{v}} = \frac{n C_{p} Δ T}{n C_{v} Δ T} = \frac{C_{p}}{C_{v}} = 1.4$
$∴ {Δ Q}_{v} = \frac{{Δ Q}_{p}}{1.4} = 50 c a l$