Question
An ideal gas changes its state from L to M by two path LNM and LM.

  1. Is the work done same for two paths?
  2. 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

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

$\text{W}_{\text{NM}}=\text{P}[\text{V}_\text{M}-\text{V}_\text{M}]=10[6-2]=40\text{J}$

$\text{W}_\text{LMN}=\text{W}_{\text{NM}}=0+40=40\text{J}$

Along LM WLM = Area under the curve LM

= Area of $\Delta\text{LMQ}+$ Area of rectangle 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.

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

$\Delta\text{Q}=400\text{J}$

$\text{dQ}=\text{dU}+\text{dW}$

$=(\text{U}_\text{M}-\text{U}_\text{L})+\Delta\text{W}_{\text{LM}}$

$\text{U}_\text{M}=\text{dQ}+\text{U}_\text{L}-\Delta\text{W}_\text{LM}$

$=400+20-30$

$=390\text{J}$

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