1. Tardigrade
  2. Question
  3. Chemistry
  4. When 0.1 mole of a gas absorbs 41.75 J of heat, the rise in temperature occurs equal to 20°C. The gas must be

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. 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

Thermodynamics

Solution:

Atomicity of gas is deiced by the ratio $C_{P} / C_{V}$. $q=n \dot{C}_{V} \Delta T$

Thus molar heat capacity at constant volume $C_{V}=\frac{q}{n \Delta T}$

$=\frac{41.75 J}{(0.1 mol )(20 K )}$

$20.88 JK ^{-1} mol ^{-1}$

For an ideal gas, $C_{P}-C_{V}=R$

or $C_{P}=R+C_{V}$

$=(8.314)+(20.88)$

$=29.19 J K^{-1} mol ^{-1}$

Finally, $\frac{C_{P}}{C_{V}}=\frac{29.19}{20.88}=1.40$

This implies than the gas is diatomic.