7.5 dm3 of an unknown gas at STP requires 110 J of heat to raise its t

1.	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 Cv , Cp and atomicity of the gas.
Answer» \(\because\) 22.4 dm³of 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.

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 Cv , Cp and atomicity of the gas.

Answer»

\(\because\) 22.4 dm³of 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.

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