Answer to Question #306299 in Physics for Phil

Question

1270.cm of propane at 18.0°C are heated and allowed to expand to 1530.cm through an isobaric process. Calculate the final temperature in °C.

Answer 1

According to the Charles's law:

$$V_1T_2 = V_2T_1$$

where $V_1 = 1270cm^3, V_2 = 1530cm^3$ are initial and final volumes respectively, and $T_1 = 18\degree C = 291K, T_2$ are initial and final temperatures respectively. Expressing $T_2$, obtain:

$$T_2 = \dfrac{V_2T_1}{V_1} = \dfrac{1530cm^3\cdot 291K}{1270cm^3} \approx 351K = 78.0\degree C$$

Answer. $78.0\degree C$.