A fixed quantity of a gas at 25 ° C, occupies a volume of 5.00 L at 1.00 atm. What is the final temperature of this same gas if the pressure is increased to 2.00 atm and the volume is doubled to 10.0 L?

Question

Quizplus · Accepted Answer

The final temperature can be found using the combined gas law: $P_1V_1/T_1 = P_2V_2/T_2$. Given $P_1 = 1.00$ atm, $V_1 = 5.00$ L, $T_1 = 25 °C = 298 K$, $P_2 = 2.00$ atm, and $V_2 = 10.0$ L, we solve for $T_2$. Rearranging the formula gives $T_2 = P_2V_2T_1/(P_1V_1)$. Substituting the given values, $T_2 = (2.00 \times 10.0 \times 298) / (1.00 \times 5.00) = 5960 / 5 = 1192 K$. Converting back to Celsius, $T_2 = 1192 - 273 = 919 °C$.

A Fixed Quantity of a Gas at 25 ° C