For a monoatomic gas- if the specific heat for a process is C-5 R -2- then density v - s temperature graph for the process will be-

The correct option is D For monoatomic C_P = 5R/2 If C = 5R/2 Then C = C_P, process is isobaric (P is constant) P = n/VRT = m/MV RT = ρ RT/M ρ T = PM/R = const ...

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Standard XII

Chemistry

Relation between Cp and Cv

For a monoato...

Question

For a monoatomic gas, if the specific heat for a process is $C = \frac{5 R}{2},$ then density v/s temperature graph for the process will be:

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Solution

The correct option is D
For monoatomic $C_{P} = \frac{5 R}{2}$
If C = $\frac{5 R}{2}$
Then $C = C_{P}$ , process is isobaric (P is constant)
$P = \frac{n}{V} R T = \frac{m}{M V} R T = \frac{ρ R T}{M}$
$ρ T = \frac{P M}{R} = c o n s t$
(Hyperbolic nature)

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For a monoatomic gas, if the specific heat for a process is C=5R2, then density v/s temperature graph for the process will be:

The correct option is D For monoatomic CP=5R2 If C = 5R2 Then C=CP, process is isobaric (P is constant) P=nVRT=mMVRT =ρRTM ρT=PMR=const (Hyperbolic nature)

For a monoatomic gas, if the specific heat for a process is $C = \frac{5 R}{2},$ then density v/s temperature graph for the process will be:

The correct option is D
For monoatomic $C_{P} = \frac{5 R}{2}$
If C = $\frac{5 R}{2}$
Then $C = C_{P}$ , process is isobaric (P is constant)
$P = \frac{n}{V} R T = \frac{m}{M V} R T = \frac{ρ R T}{M}$
$ρ T = \frac{P M}{R} = c o n s t$
(Hyperbolic nature)