JAMB - Physics (2024 - No. 98)

If the pressure on an enclosed gas is doubled and its temperature rises from 27ºC to 127ºC, then its volume will 
increase by \(\frac{2}{3^{rd}}\)
increase by \(\frac{1}{3^{rd}}\)
decrease by \(\frac{2}{3^{rd}}\)
decrease by \(\frac{1}{3^{rd}}\)

Explanation

Using the general gas equation;

\(\frac{P_1 \times V_1}{T_1}\) = \(\frac{P_2 \times V_2}{T_2}\)

P\(_1\) = P, P\(_2\) = 2P, T\(_1\) = 27ºC → 300K, T\(_2\) = 127ºC → 400k,

V\(_2\) = \(\frac{P_1 \times V_1 \times T_2}{T_1 \times P_2}\)

V\(_2\) =  \(\frac{P \times V_1 \times 400}{300 \times 2P}\) = \(\frac{2V_1}{3}\)

ΔV = V\(_1\) - \(\frac{2V_1}{3}\) = \(\frac{V_1}{3}\)

Therefore, The volume decreases by \(\frac{1}{3}\) of the original volume.

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