A perfect gas (0.1 mol) having overline{C}_{ | vedclass

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Question

(B) The process from $1$ to $4$ consists of three steps: $1 \rightarrow 2$,$2 \rightarrow 3$,and $3 \rightarrow 4$.
Step $1 \rightarrow 2$: Isochoric

Vedclass · Accepted Answer

$304$

Vedclass · Answer

(B) The process from $1$ to $4$ consists of three steps: $1 ightarrow 2$,$2 ightarrow 3$,and $3 ightarrow 4$.
Step $1 ightarrow 2$: Isochoric process $(V = 1000 \ cm^3 = 1 \ L)$,so $W_{1 ightarrow 2} = 0 \ J$.
Step $2 ightarrow 3$: Isobaric expansion $(P = 3 \ atm)$ from $V_2 = 1 \ L$ to $V_3 = 2 \ L$.
$W_{2 ightarrow 3} = -P \Delta V = -3 \ atm 	imes (2 \ L - 1 \ L) = -3 \ L \ atm$.
Step $3 ightarrow 4$: Isochoric process $(V = 2000 \ cm^3 = 2 \ L)$,so $W_{3 ightarrow 4} = 0 \ J$.
Total work done $W = W_{1 	o 2} + W_{2 	o 3} + W_{3 	o 4} = 0 + (-3 	ext{ L atm}) + 0 = -3 	ext{ L atm}$
Converting to Joules: $W = -3 	imes 101.3 \ J = -303.9 \ J$.
Rounding to the nearest integer,the magnitude is $304 \ J$.

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