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The temperature of two moles of an ideal diatomic gas is decreased from 𝟔𝟎𝟎 𝐊 to 𝟑𝟎𝟎 𝐊 at constant pressure. Is entropy change negative?

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    $\begingroup$ The entropy change of a system can be negative. However the total change in entropy (system + surroundings) is always non-negative. $\endgroup$ – Internet Guy Jun 10 at 4:49
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Entropy change of the gas system can indeed be negative. You can determine it by using the following equation for change in entropy as a function of change on temperature and volume: $$\Delta S (T,V) = C_V \ln(\frac{T_2}{T_1}) + R \ln (\frac{V_2}{V_1})$$ where $C_V$ is the heat capacity at constant volume, and $R$ is the ideal gas constant. The fact that your gas is diatomic goes in the value of $C_V$. The only element for the equation above that you are missing is the final volume, which you can easily calculate from the ideal gas law evaluated at the initial and final configurations: $$PV=nRT$$

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You must keep in mind that entropy change of a system can be negative positive or zero. But for the case of a irreversible process the change in entropy of system and surrounding has to be non-negative. As the pressure is constant in above question and temperature it self is decreasing son volume will also has to be lowered. And as you know by the relation of entropy $$ Δ𝑆(𝑇,𝑉)=𝐶𝑉 ln{(𝑇_2/𝑇_1)}+𝑅 ln(𝑉_2/𝑉_1) $$

Here we can easily see that for the given condition i.e. $$ T_1>T_2 ; V_1>V_2 $$ We’ll have natural logs negative and so entropy will also be negative. Hence concluding that change in entropy of gas will be negative.

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