The figure below shows two paths that may be | vedclass

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(B) From the first law of thermodynamics,$\Delta U = Q - W$. Since the change in internal energy $\Delta U$ is a state function,it is the same for any

Vedclass · Accepted Answer

$460$

Vedclass · Answer

(B) From the first law of thermodynamics,$\Delta U = Q - W$. Since the change in internal energy $\Delta U$ is a state function,it is the same for any path between two states.For path $ABC$: $\Delta U_{ABC} = \Delta U_{AC} = Q_{ABC} - W_{ABC}$.Total heat absorbed $Q_{ABC} = Q_{AB} + Q_{BC} = 400 \, J + 100 \, J = 500 \, J$.The work done $W_{ABC}$ is the area under the path $ABC$ on the $PV$ diagram. Since $AB$ is an isochoric process,$W_{AB} = 0$. For process $BC$,$W_{BC} = P \Delta V = (6 	imes 10^4 \, Pa) 	imes (4 	imes 10^{-3} \, m^3 - 2 	imes 10^{-3} \, m^3) = 6 	imes 10^4 	imes 2 	imes 10^{-3} = 120 \, J$.So,$W_{ABC} = 0 + 120 \, J = 120 \, J$.Thus,$\Delta U_{AC} = 500 \, J - 120 \, J = 380 \, J$.For the process $AC$,$\Delta U_{AC} = Q_{AC} - W_{AC}$.The work done $W_{AC}$ is the area under the line $AC$,which is a trapezoid: $W_{AC} = \frac{1}{2} 	imes (P_A + P_C) 	imes (V_C - V_A) = \frac{1}{2} 	imes (2 	imes 10^4 + 6 	imes 10^4) 	imes (4 	imes 10^{-3} - 2 	imes 10^{-3}) = \frac{1}{2} 	imes 8 	imes 10^4 	imes 2 	imes 10^{-3} = 80 \, J$.Therefore,$Q_{AC} = \Delta U_{AC} + W_{AC} = 380 \, J + 80 \, J = 460 \, J$.

Molecule type	$\frac{C_P}{C_V}$
$A$. Monoatomic	$I$. $\frac{7}{5}$
$B$. Diatomic rigid molecules	$II$. $\frac{9}{7}$
$C$. Diatomic non-rigid molecules	$III$. $\frac{4}{3}$
$D$. Triatomic rigid molecules	$IV$. $\frac{5}{3}$

The figure below shows two paths that may be taken by a gas to go from a state $A$ to a state $C$. In process $AB$,$400 \, J$ of heat is added to the system and in process $BC$,$100 \, J$ of heat is added to the system. The heat absorbed by the system in the process $AC$ will be ............. $J$.

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