An ideal gas changes its state from L to M by two path LNM and LM… - Vidyadip

Question

An ideal gas changes its state from $L$ to $M$ by two path $\text{LNM}$ and $LM$.

Is the work done same for two paths?
The internal energy of gas at $L$ is $20J$ and the amount of heat needed to change its state through $LM$ is $400J$. What is the internal energy of gas at $M$?

✓

Answer

$\text{W}_{\text{LN}}=\text{PdV}=0$

$\text{W}_{\text{NM}}=\text{P}[\text{V}_\text{M}-\text{V}_\text{M}]=10[6-2]=40\text{J}$
$\text{W}_\text{LMN}=\text{W}_{\text{NM}}=0+40=40\text{J}$
Along $\ce{LM \ W_{LM}} =$ Area under the curve $LM$
= Area of $\Delta\text{LMQ}+$ Area of rectangle $\text{LQZP.}$
$=\frac12\times\text{LQ}\times\text{MQ}+\text{LP}\times\text{PZ}$
$=\frac12\times4\times5+5\times4$
$=10+20=30\text{J}$
So work done is less along $LM.$

$\text{U}_\text{L}=20\text{J}$

$\Delta\text{Q}=400\text{J}$
$\text{dQ}=\text{dU}+\text{dW}$
$=(\text{U}_\text{M}-\text{U}_\text{L})+\Delta\text{W}_{\text{LM}}$
$\text{U}_\text{M}=\text{dQ}+\text{U}_\text{L}-\Delta\text{W}_\text{LM}$
$=400+20-30$
$=390\text{J}$

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