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

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 (A) triatomic (B) diatomic (C) polyatomic (D) monoatomic

Tardigrade Admin · Accepted Answer

Atomicity of gas is deiced by the ratio

C_{P} / C_{V}

.

q = n \dot{C}_{V} Δ T

Thus molar heat capacity at constant volume

C_{V} = \frac{q}{n Δ T}

= \frac{41.75 J}{( 0.1 m o l ) ( 20 K )}

20.88 J K^{- 1} m o l^{- 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} m o l^{- 1}

Finally,

\frac{C _{P}}{C _{V}} = \frac{29.19}{20.88} = 1.40

This implies than the gas is diatomic.

Q. 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

triatomic

diatomic

polyatomic

monoatomic

Q. When 0.1 mole of a gas absorbs 41.75 J of heat, the rise in temperature occurs equal to $2 0^{\circ}$ C. The gas must be