Question

7.5 dm³ of an unknown gas at STP requires 110 J of heat to raise its temperature by 15°C at constant volume. Calculate C_v , C_p and atomicity of the gas.

Answer 1

\(\because\) 22.4 dm³ of gas at STP = 1mol

\(\therefore\) 7.5 dm³ of a gas STP = \(\frac{1}{22.4}\)x 7.5

= 0.33 mol

\(\because\) For 15°C rice, 0.33 moles of the gas at constant volume requires heat = 110 J

\(\therefore\) For 1°C rise, 1 mole of the gas at constant volume will require heat

= \(\frac{110}{15\times0.33}\)

= \(\frac{110}{4.95}\) = 22.222 J

\(\therefore\) C_v = 22.222 JK⁻¹ mol⁻¹ = 22.22 J K⁻¹mol⁻¹

\(\because\) C_p − C_v = R

or C_p = C_v + R

= 22.222 J K⁻¹ + 8.314 J K⁻¹ mol⁻¹

= 30.536 = 30.53 J K⁻¹ mol⁻¹

y = \(\frac{C_p}{C_v}\) = \(\frac{30.53}{22.22}\) = 1.37

\(\therefore\) The gas is triatomic.