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 \(\frac{1}{3^{rd}}\)
The correct answer is: D
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.
There is an explanation video available .