0.08 text{ kg} of air is heated at constant v | vedclass

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(C) For a process at constant volume,the work done $W = 0$. According to the first law of thermodynamics,$\Delta Q = \Delta U + W$,so $\Delta Q = \Del

Vedclass · Accepted Answer

$284$

Vedclass · Answer

(C) For a process at constant volume,the work done $W = 0$. According to the first law of thermodynamics,$\Delta Q = \Delta U + W$,so $\Delta Q = \Delta U$.The change in internal energy is given by $\Delta U = m \cdot c_v \cdot \Delta T$.Given:$m = 0.08 	ext{ kg}$$c_v = 0.17 	ext{ kcal/kg}^{\circ} 	ext{C} = 0.17 	imes 1000 	ext{ cal/kg}^{\circ} 	ext{C} = 170 	ext{ cal/kg}^{\circ} 	ext{C}$$\Delta T = 5^{\circ} 	ext{C}$$J = 4.18 	ext{ J/cal}$Substituting the values:$\Delta U = 0.08 	ext{ kg} 	imes 170 	ext{ cal/kg}^{\circ} 	ext{C} 	imes 5^{\circ} 	ext{C} 	imes 4.18 	ext{ J/cal}$$\Delta U = 0.08 	imes 170 	imes 5 	imes 4.18 	ext{ J}$$\Delta U = 68 	imes 4.18 	ext{ J}$$\Delta U = 284.24 	ext{ J}$Thus,the change in internal energy is approximately $284 	ext{ J}$.

Column-$I$	Column-$II$
$(a)$ Horizontal line parallel to $V$-axis	$(i)$ Adiabatic process
$(b)$ Vertical line parallel to $P$-axis	$(ii)$ Isobaric process
	$(iii)$ Isochoric process

$0.08 \text{ kg}$ of air is heated at constant volume through $5^{\circ} \text{C}$. The specific heat of air at constant volume is $0.17 \text{ kcal/kg}^{\circ} \text{C}$ and $J = 4.18 \text{ J/cal}$. The change in its internal energy is approximately: (in $\text{ J}$)

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Match the graphs in Column-$I$ with the processes in Column-$II$:
Column-$I$ Column-$II$
$(a)$ Horizontal line parallel to $V$-axis $(i)$ Adiabatic process
$(b)$ Vertical line parallel to $P$-axis $(ii)$ Isobaric process
$(iii)$ Isochoric process

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