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# When 0.1 mole of a gas absorbs 41.75 J of heat, the rise in temperature occurs equal to $20 ^\circ C$. The gas must be

(a) triatomic
(b) diatomic
(c) polyatomic
(d) monoatomic

Toolbox:
• $C_P = C_V + R$
$C_V$ (heat absorbed per degree rise per mole) = $\frac{41.75J}{0.1 mol \times 20^\circ C} = 20.875 JK^{-1}mol^{-1}$
$C_P = C_V + R = 20.875 + 8.314 KJ^{-1}mol^{-1} = 29.189 JK^{-1}mol^{-1}$
$\frac{C_P}{C_V} = \frac{29.189}{20.875} = 1.40$
$\therefore$ The gas is monoatomic.