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Standard XII

Physics

Thermodynamic Processes

Figure shows ...

Question

Figure shows two processes a and b for a given sample of a gas> If $△ Q_{1}, △ Q_{2}$ are the amounts of heat absorbed by the system in the cases and $△ U_{1}, △ U_{2}$ are changes in internal energies respectively, then

△ Q_{1} = △ Q_{2}; △ U_{1} = △ U_{2}

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△ Q_{1} > △ Q_{2}; △ U_{1} > △ U_{2}

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△ Q_{1} < △ Q_{2}; △ U_{1} < △ U_{2}

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△ Q_{1} > △ Q_{2}; △ U_{1} = △ U_{2}

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Solution

The correct option is D $△ Q_{1} > △ Q_{2}; △ U_{1} = △ U_{2}$
Internal energy is a state function.
$∴ Δ U_{a} = Δ U_{b}$
Work done along a is greater than that along b.
$W_{a} > W_{b}$
$Δ Q_{a} = Δ U_{a} + W_{a}$
$Δ Q_{b} = Δ U_{b} + W_{b}$
$∴ Δ Q_{a} > Δ Q_{b}$

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Figure shows two processes a and b for a given sample of a gas> If △Q1,△Q2 are the amounts of heat absorbed by the system in the cases and△U1,△U2 are changes in internal energies respectively, then

The correct option is D △Q1>△Q2;△U1=△U2Internal energy is a state function.∴ΔUa=ΔUbWork done along a is greater than that along b.Wa>WbΔQa=ΔUa+WaΔQb=ΔUb+Wb∴ΔQa>ΔQb

Figure shows two processes a and b for a given sample of a gas> If $△ Q_{1}, △ Q_{2}$ are the amounts of heat absorbed by the system in the cases and $△ U_{1}, △ U_{2}$ are changes in internal energies respectively, then

The correct option is D $△ Q_{1} > △ Q_{2}; △ U_{1} = △ U_{2}$
Internal energy is a state function.
$∴ Δ U_{a} = Δ U_{b}$
Work done along a is greater than that along b.
$W_{a} > W_{b}$
$Δ Q_{a} = Δ U_{a} + W_{a}$
$Δ Q_{b} = Δ U_{b} + W_{b}$
$∴ Δ Q_{a} > Δ Q_{b}$