JEE MAIN - Chemistry (2023 - 24th January Evening Shift - No. 16)
One mole of an ideal monoatomic gas is subjected to changes as shown in the graph. The magnitude of the work done (by the system or on the system) is __________ J (nearest integer)
_24th_January_Evening_Shift_en_16_1.png)
Given ; $$\log2=0.3$$
$$\ln10=2.3$$
Answer
620
Explanation
$1 \rightarrow 2 \Rightarrow$ Isobaric process
$2 \rightarrow 3 \Rightarrow$ Isochoric process
$3 \rightarrow 1 \Rightarrow$ Isothermal process
$\mathrm{W}=\mathrm{W}_{1 \rightarrow 2}+\mathrm{W}_{2 \rightarrow 3}+\mathrm{W}_{3 \rightarrow 1}$
$=\left(-\mathrm{P}\left(\mathrm{V}_2-\mathrm{V}_1\right)+0\left[-\mathrm{P}_1 \mathrm{~V}_1 \ln \left(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\right)\right]\right)$
$=\left[-1 \times(40-20)+0+\left[-1 \times 20 \ln \left(\frac{20}{40}\right)\right]\right]$
$=-20+20 \ln 2$
$=-20+20 \times 2.3 \times 0.3$
$=-6.2$ bar $L$
$|\mathrm{W}|=6.2$ bar $1=620 \mathrm{~J}$
$2 \rightarrow 3 \Rightarrow$ Isochoric process
$3 \rightarrow 1 \Rightarrow$ Isothermal process
$\mathrm{W}=\mathrm{W}_{1 \rightarrow 2}+\mathrm{W}_{2 \rightarrow 3}+\mathrm{W}_{3 \rightarrow 1}$
$=\left(-\mathrm{P}\left(\mathrm{V}_2-\mathrm{V}_1\right)+0\left[-\mathrm{P}_1 \mathrm{~V}_1 \ln \left(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\right)\right]\right)$
$=\left[-1 \times(40-20)+0+\left[-1 \times 20 \ln \left(\frac{20}{40}\right)\right]\right]$
$=-20+20 \ln 2$
$=-20+20 \times 2.3 \times 0.3$
$=-6.2$ bar $L$
$|\mathrm{W}|=6.2$ bar $1=620 \mathrm{~J}$
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