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In the given figure, an ideal gas changes its state from $\mathrm{A}$ to state $\mathrm{C}$ by two paths $\mathrm{\mathrm{ABC} \: and\: \mathrm{AC}}$ .The internal energy of gas at $\mathrm{A}$ is $10 \mathrm{~J}$ and amount of heat supplied to change its state to $\mathrm{C}$ through the path $\mathrm{AC}$ is $200 \mathrm{~J}$ . Calculate the internal energy at $\mathrm{C}$ .

Option: 1
$140 \mathrm{~J}$

Option: 2
$150 \mathrm{~J}$

Option: 3
$160 \mathrm{~J}$

Option: 4
$170 \mathrm{~J}$

Answers (1)

best_answer

Since the work done $\mathrm{W}=\int P d V=$ area under $\mathrm{P}-\mathrm{V}$ curve, so

$\mathrm{ \mathrm{W}_{\mathrm{ABC}}=\mathrm{W}_{\mathrm{AB}}+\mathrm{W}_{\mathrm{BC}} }$

$\mathrm{ \text { i.e. } \quad \mathrm{W}_{\mathrm{ABC}}=0+15 \times 4=60 \mathrm{~J} }$

$\mathrm{ \text { and } \quad \mathrm{W}_{\mathrm{AC}}=\frac{1}{2}(5+15) \times(6-2)=40 \mathrm{~J} }$

thus the work done along $\mathrm{A C}$ is least.

According to first law of thermodynamics,

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

so for path $\mathrm{ \mathrm{AC}, }$

$\mathrm{ \left(\mathrm{U}_{\mathrm{C}}-\mathrm{U}_{\mathrm{A}}\right)=\mathrm{dQ}-\mathrm{dW}=200-40=160 \mathrm{~J} }$

$\mathrm{ so \quad \mathrm{U}_{\mathrm{C}}=160+\mathrm{U}_{\mathrm{A}}=160+10=170 \mathrm{~J} } (\because U_A=10)$

Posted by

Suraj Bhandari

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