Question 735: Thermodynamics & KTG

Question

A gas expands in such a way that its pressure and volume satisfy the condition $\mathrm{PV}^2=$ constant. Then the temperature of the gas

Accepted Answer

To analyze how the temperature of a gas changes during expansion, given that the pressure $ P $ and volume $ V $ satisfy the condition $ \mathrm{PV}^2 = 	ext{constant} $, we start with the ideal gas equation:
$ \mathrm{PV} = \mathrm{nRT} $
Using the given condition:
$ \mathrm{PV}^2 = 	ext{constant} $
From the ideal gas equation, we can express $ P $ as:
$ P = \frac{nRT}{V} $
Substituting this into the given condition:
$ \left(\frac{nRT}{V}ight) V^2 = 	ext{constant} $
This simplifies to:
$ nRT \cdot V = 	ext{constant} $
Therefore, we find that:
$ TV = 	ext{constant} $
From this relationship, we observe that:
$ T \propto \frac{1}{V} $
Thus, if the gas expands (meaning $ V $ increases), the temperature $ T $ must decrease to maintain the equation $ TV = 	ext{constant} $. Therefore, during expansion under this condition, the temperature of the gas decreases.

Answer

will decrease.

Answer

will increase.

Answer

will not change.

Answer

may increase or decrease depending upon the values of pressure and volume.

Thermodynamics & KTG

A gas expands in such a way that its pressure and volume satisfy the condition \mathrm{PV}^2= constant. Then the temperature of the gas

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