A 16.4-L sample of gas at STP is heated to 55°C at 605 torr. What is the new volume?
A) 24.8 L
B) 4.15 L
C) 17.1 L
D) 15.7 L
E) 16.6 L

Question

Quizplus · Accepted Answer

To solve this problem, we use the combined gas law, which is $P_1V_1/T_1 = P_2V_2/T_2$. Given that the initial conditions are at STP (Standard Temperature and Pressure), we know that $P_1 = 760$ torr, $V_1 = 16.4$ L, and $T_1 = 273$ K (0°C). The final conditions are $P_2 = 605$ torr, $T_2 = 328$ K (55°C + 273). We need to find $V_2$.Rearranging the combined gas law to solve for $V_2$ gives us $V_2 = V_1 \times (P_1/P_2) \times (T_2/T_1)$.Plugging in the values, we get $V_2 = 16.4 \times (760/605) \times (328/273)$.Calculating this gives a $V_2$ of approximately 24.8 L.

A 164-L Sample of Gas at STP Is Heated to 55°C