When 2 text{ moles} of a monatomic gas expand | vedclass

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(C) For an adiabatic process,the work done is given by $W = -\Delta U = -\frac{f}{2} nR \Delta T = \frac{f}{2} nR (T_i - T_f)$.For the monatomic gas:

Vedclass · Accepted Answer

$2.5$

Vedclass · Answer

(C) For an adiabatic process,the work done is given by $W = -\Delta U = -\frac{f}{2} nR \Delta T = \frac{f}{2} nR (T_i - T_f)$.For the monatomic gas: $n_1 = 2$,$f_1 = 3$,$T_{i1} = 80^{\circ} C$,$T_{f1} = 50^{\circ} C$.$W = \frac{3}{2} 	imes 2 	imes R 	imes (80 - 50) = 3R 	imes 30 = 90R$.For the diatomic gas: $n_2 = 3$,$f_2 = 5$,$T_{i2} = 50^{\circ} C$,$T_{f2} = 20^{\circ} C$.$W' = \frac{5}{2} 	imes 3 	imes R 	imes (50 - 20) = \frac{15}{2} R 	imes 30 = 15R 	imes 15 = 225R$.Now,find the ratio: $\frac{W'}{W} = \frac{225R}{90R} = \frac{225}{90} = 2.5$.Therefore,$W' = 2.5 W$.

When $2 \text{ moles}$ of a monatomic gas expands adiabatically from a temperature of $80^{\circ} C$ to $50^{\circ} C$,the work done is $W$. The work done when $3 \text{ moles}$ of a diatomic gas expands adiabatically from $50^{\circ} C$ to $20^{\circ} C$ is: (in $W$)

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