A gas is cooled at a constant pressure from 57ºC was observed to shrink one-fifth (1\5) of its original volume of 2.00cm\(^3\). Find its new temperature

Answer Details

At constant pressure a fixed mass of gas obeys Charles' law: the volume is directly proportional to the absolute temperature, so

\[\frac{V_1}{T_1} = \frac{V_2}{T_2}, \qquad T\ \text{in kelvin}.\]

Converting the initial temperature to kelvin is the essential first step, because a ratio of Celsius temperatures is meaningless:

\[T_1 = 57 + 273 = 330\,\mathrm{K}, \qquad V_1 = 2.00\,\mathrm{cm^{3}}.\]

The gas shrinks to one-fifth of its original volume, so \(V_2 = \tfrac{1}{5}\times 2.00 = 0.40\,\mathrm{cm^{3}}\). Rearranging Charles' law:

\[T_2 = T_1\times\frac{V_2}{V_1} = 330 \times \frac{0.40}{2.00} = 330 \times \frac{1}{5} = 66\,\mathrm{K}.\]

Converting back to the Celsius scale asked for in the options:

\[\theta_2 = 66 - 273 = -207\,^{\circ}\mathrm{C}.\]

Notice how the wording controls the arithmetic. Read as "the volume becomes one-fifth of the original", the volume ratio is \(1/5\) and the temperature falls by the same factor, giving \(-207\,^{\circ}\mathrm{C}\). Read instead as "the volume falls by one-fifth", the ratio would be \(4/5\) and the answer would be \(330\times0.8 = 264\,\mathrm{K} = -9\,^{\circ}\mathrm{C}\), which is not offered, so the first reading is the intended one.

The commonest error in this topic is to work in degrees Celsius, which here would give \(57/5 \approx 11\,^{\circ}\mathrm{C}\) and is completely wrong because the gas laws are proportionalities measured from absolute zero, not from the ice point. Always convert to kelvin before forming any ratio, and convert back only at the last line.