An ideal gas undergoes the process $1 ⟶ 2$ as shown in the figure, the heat supplied and work done in process is $Δ Q$ and $Δ W$ respectively. the ratio $Δ Q : Δ W$ is

γ : γ - 1

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γ

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γ - 1

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γ - 1 : γ

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Solution

The correct option is D $γ : γ - 1$
$Expalantion:-$

We know that,

$P V = n R T$

$\Rightarrow \frac{V}{T} = \frac{n R}{P} = c o n s t a n t$

$\Rightarrow$ Pressure constant. It is an isobaric process

$Δ Q = Δ U + Δ W$

$\frac{Δ Q}{Δ W} = \frac{n C_{P} d T}{Δ Q - Δ U}$

$= \frac{n C_{P} d T}{n C_{P} d T - n C_{V} d T}$

$= \frac{C_{P}}{C_{P} - C_{V}}$

$= \frac{\frac{C_{P}}{C_{V}}}{\frac{C_{P}}{C_{V}} - 1}$

$= \frac{γ}{γ - 1}$

$Hence the correct option is A$

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An ideal gas undergoes the process 1⟶2 as shown in the figure, the heat supplied and work done in process is ΔQ and ΔW respectively. the ratio ΔQ:ΔW is

The correct option is D γ:γ−1Expalantion:-We know that,PV=nRT⇒VT=nRP=constant⇒ Pressure constant. It is an isobaric process ΔQ=ΔU+ΔWΔQΔW=nCPdTΔQ−ΔU=nCPdTnCPdT−nCVdT=CPCP−CV=CPCVCPCV−1=γγ−1Hence the correct option is A

An ideal gas undergoes the process $1 ⟶ 2$ as shown in the figure, the heat supplied and work done in process is $Δ Q$ and $Δ W$ respectively. the ratio $Δ Q : Δ W$ is