The average degree of freedom per molecule for a gas is 6. The gas performs 25 J of work when it expands at constant pressure. The heat absorbed by the gas is

Question

The average degree of freedom per molecule for a gas is 6. The gas performs 25 J of work when it expands at constant pressure. The heat absorbed by the gas is (A) 75 J (B) 100 J (C) 150 J (D) 125 J

Tardigrade Admin · Accepted Answer

&gamma; = 1 + (2/f) = 1 + (2/6) = (4/3) (Δ W/Δ Q)=(Δ Q - Δ U/Δ Q) = 1-(nCV Δ T/nCP Δ T) = 1 - (1/&gamma;) = 1 -(3/4) =(1/4) ∴ : : Δ Q = 4 Δ W = 100 J ( ∵ (CP/CV) = &gamma;)

Q. The average degree of freedom per molecule for a gas is $6$ . The gas performs $25 J$ of work when it expands at constant pressure. The heat absorbed by the gas is

75 J

100 J

150 J

125 J

$γ = 1 + \frac{2}{f} = 1 + \frac{2}{6} = \frac{4}{3}$
$\frac{Δ W}{Δ Q} = \frac{Δ Q - Δ U}{Δ Q} = 1 - \frac{n C _{V} Δ T}{n C _{P} Δ T}$
$= 1 - \frac{1}{γ} = 1 - \frac{3}{4} = \frac{1}{4}$
$∴ Δ Q = 4Δ W = 100 J$
$(∵ \frac{C _{P}}{C _{V}} = γ)$

Q. The average degree of freedom per molecule for a gas is 6. The gas performs 25J of work when it expands at constant pressure. The heat absorbed by the gas is