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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 equal to unity. The molar heat capacity of gas is:

$(a)\;2.0R\qquad(b)\;1.5R\qquad(c)\;2.5R\qquad(d)\;0$

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A)
Let P , V be the pressure and Volume of gas at temperature T
$P_1V_1 =RT$
$P_2(V_1+dv) = R(T + 1)$
$\therefore P_2 = RT+R$
Since ($\large\frac{P_2}{V_1 + dv} = 1$)
$\Rightarrow 2(\large\frac{\delta P_2}{\delta T})_v = R$
$\Rightarrow (\large\frac{\delta P_2}{\delta T}) = \large\frac{R}{2}$
Since, $C = C_v + \large\frac{\delta P}{\delta T} $
$= \large\frac{3R}{2} + \large\frac{R}{2} $
= 2R
Hence answer is (a)
Pls tell that how Cv is 3R /2
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