In the equation PV- - constant- the value of - is unity- Then the process is

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Standard XII

Physics

Thermodynamic Processes

In the equati...

Question

In the equation $P V^{γ}$ = constant, the value of $γ$ is unity. Then the process is

isothermal

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adiabatic

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isobaric

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irreversible

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Solution

The correct option is B isothermal
Given that:
$P V^{γ} = c o n s t a n t$
Given that $γ = 1$
$P V = c o n s t a n t$
For an ideal gas:
$P V = n R T = c o n s t a n t$
i.e.
$n R T = c o n s t a n t$
$T = \frac{c o n s t a n t}{n R} = c o n s t a n t$
Hence the process is isothermal.

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In the equation PVγ = constant, the value of γ is unity. Then the process is

The correct option is B isothermalGiven that:PVγ=constantGiven that γ=1PV=constantFor an ideal gas:PV=nRT=constanti.e. nRT=constantT=constantnR=constantHence the process is isothermal.

In the equation $P V^{γ}$ = constant, the value of $γ$ is unity. Then the process is

The correct option is B isothermal
Given that:
$P V^{γ} = c o n s t a n t$
Given that $γ = 1$
$P V = c o n s t a n t$
For an ideal gas:
$P V = n R T = c o n s t a n t$
i.e.
$n R T = c o n s t a n t$
$T = \frac{c o n s t a n t}{n R} = c o n s t a n t$
Hence the process is isothermal.