A monoatomic ideal gas undergoes a process in which the ratio of $P$ to $V$ at any instant is constant and equals to $1$ . What is the molar heat capacity of the gas?

\frac{4 R}{2}

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\frac{3 R}{2}

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\frac{5 R}{2}

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Solution

The correct option is A $\frac{4 R}{2}$
$P = V$ given, $\frac{P}{V} = 1$
$d g = c_{v} d t + P d v$
For $1$ mole of gas $P V = R t$
$P d v + V d P = R d T 2 P d V = R d T P d V = \frac{R d T}{2}$
$d g = C_{r} d t + \frac{R d T}{2}$
$= \frac{3}{2} R + \frac{R}{2}$
$= \frac{4 R}{2}$

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A monoatomic ideal gas undergoes a process in which the ratio of P to V at any instant is constant and equals to 1. What is the molar heat capacity of the gas?

The correct option is A 4R2P=V given, PV=1dg=cvdt+PdvFor 1 mole of gas PV=RtPdv+VdP=RdT2PdV=RdTPdV=RdT2dg=Crdt+RdT2=32R+R2=4R2

A monoatomic ideal gas undergoes a process in which the ratio of $P$ to $V$ at any instant is constant and equals to $1$ . What is the molar heat capacity of the gas?

The correct option is A $\frac{4 R}{2}$
$P = V$ given, $\frac{P}{V} = 1$
$d g = c_{v} d t + P d v$
For $1$ mole of gas $P V = R t$
$P d v + V d P = R d T 2 P d V = R d T P d V = \frac{R d T}{2}$
$d g = C_{r} d t + \frac{R d T}{2}$
$= \frac{3}{2} R + \frac{R}{2}$
$= \frac{4 R}{2}$