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Standard X

Physics

Principle of Calorimetry

70 cal heat i...

Question

70 cal heat is required to raise the temperature of 2 moles of an ideal gas at constant pressure from $25^{o} C$ to $30^{o} C$ , then the amount of heat required to raise the temperature of same gas through same raise of temp. at constant volume is

50 c a l

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70 c a l

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90 c a l

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60 c a l

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Solution

The correct option is A $50 c a l$
For isobaric process, $Δ Q = n C_{p} Δ T$
Putting values, we get $\to C_{p} = 7 c a l / m o l - K$

Also, $C_{v} = C_{p} - R$
$\to C_{v} = (7 - 2) c a l / m o l - K$
Hence for isochoric poocess,
$Δ Q = n C_{v} Δ T$
Putting values, we get $\to Δ Q = 50 c a l$

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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)

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70 cal heat is required to raise the temperature of 2 moles of an ideal gas at constant pressure from 25oC to 30oC , then the amount of heat required to raise the temperature of same gas through same raise of temp. at constant volume is

The correct option is A 50calFor isobaric process, ΔQ=nCpΔTPutting values, we get →Cp=7cal/mol−K Also, Cv=Cp−R→Cv=(7−2)cal/mol−K Hence for isochoric poocess, ΔQ=nCvΔTPutting values, we get→ΔQ=50 cal

70 cal heat is required to raise the temperature of 2 moles of an ideal gas at constant pressure from $25^{o} C$ to $30^{o} C$ , then the amount of heat required to raise the temperature of same gas through same raise of temp. at constant volume is

The correct option is A $50 c a l$
For isobaric process, $Δ Q = n C_{p} Δ T$
Putting values, we get $\to C_{p} = 7 c a l / m o l - K$

Also, $C_{v} = C_{p} - R$
$\to C_{v} = (7 - 2) c a l / m o l - K$
Hence for isochoric poocess,
$Δ Q = n C_{v} Δ T$
Putting values, we get $\to Δ Q = 50 c a l$