During an experiment, an ideal gas is found to obey an additional law p2V = constant. The gas is initially at a temperature T and volume V. When it expands to a volume 2V. the temperature becomes ____ .Correct answer is '1.4'. Can you explain this answer?

Question

Answer

vp² = constant
putting ,

we have T²/V = constant
or T∝√V
so , if V is doubled , T becomes √2 = 1.4times.

Answer

Understanding the Given Law
In this experiment, the ideal gas follows the law $ p^2 V = \text{constant} $. This implies a relationship between pressure (p), volume (V), and temperature (T) of the gas.

Initial Conditions
- The initial conditions are:
- Pressure: $ p_1 $
- Volume: $ V $
- Temperature: $ T $
From the ideal gas law, we know:
$$ p_1 V = nRT $$
Where $ n $ is the number of moles and $ R $ is the gas constant.

Final Conditions After Expansion
After expanding to volume $ 2V $, let the final pressure be $ p_2 $ and the final temperature be $ T_f $:
- New volume: $ 2V $
- Applying the given law at the final state:
$$ p_2^2 (2V) = p_1^2 V $$
From this, we can derive:
$$ p_2^2 = \frac{p_1^2 V}{2V} = \frac{p_1^2}{2} $$
Thus:
$$ p_2 = \frac{p_1}{\sqrt{2}} $$

Applying the Ideal Gas Law
Now, applying the ideal gas law to the final state:
$$ p_2 (2V) = nRT_f $$
Substituting for $ p_2 $:
$$ \frac{p_1}{\sqrt{2}} (2V) = nRT_f $$
Equating with the initial state:
$$ p_1 V = nRT $$
We can solve for $ T_f $:
$$ \frac{2p_1 V}{\sqrt{2}} = nRT_f $$
Substituting $ p_1 V $:
$$ \frac{2(nRT)}{\sqrt{2}} = nRT_f $$
Simplifying gives:
$$ T_f = \frac{2T}{\sqrt{2}} = T \sqrt{2} $$
To find the numerical temperature, if $ T $ is normalized to 1, then:
$$ T_f = 1 \cdot \sqrt{2} \approx 1.4 $$

Conclusion
Hence, when the gas expands to volume $ 2V $, the final temperature becomes approximately 1.4.

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