19. a container containing 5.2 l of a gas is collected at 291.2 k and then allowed to expand to 15.1 l. what…

19. a container containing 5.2 l of a gas is collected at 291.2 k and then allowed to expand to 15.1 l. what must the new temperature be, in k, in order to maintain the same pressure? round to 1 decimal place.\n20. calculate the temperature, in k, when 94.5 l at 305.6 k is compressed to 4.7 l. round to 1 decimal place.

19. a container containing 5.2 l of a gas is collected at 291.2 k and then allowed to expand to 15.1 l. what must the new temperature be, in k, in order to maintain the same pressure? round to 1 decimal place.\n20. calculate the temperature, in k, when 94.5 l at 305.6 k is compressed to 4.7 l. round to 1 decimal place.

Answer

Explanation:

Step1: Identify the gas - law formula

For a gas at constant pressure, Charles's law is used: $\frac{V_1}{T_1}=\frac{V_2}{T_2}$, where $V_1$ and $T_1$ are the initial volume and temperature, and $V_2$ and $T_2$ are the final volume and temperature.

Step2: Solve for the new temperature in the first problem

We are given $V_1 = 5.2L$, $T_1=291.2K$, and $V_2 = 15.1L$. From $\frac{V_1}{T_1}=\frac{V_2}{T_2}$, we can re - arrange to get $T_2=\frac{V_2T_1}{V_1}$. Substitute the values: $T_2=\frac{15.1\times291.2}{5.2}$. $T_2=\frac{4397.12}{5.2}=845.6K$.

Step3: Solve for the new temperature in the second problem

We are given $V_1 = 94.5L$, $T_1 = 305.6K$, and $V_2=4L$. Using $T_2=\frac{V_2T_1}{V_1}$, substitute the values: $T_2=\frac{4\times305.6}{94.5}$. $T_2=\frac{1222.4}{94.5}\approx12.9K$.

Answer:

For the first problem: 845.6K For the second problem: 12.9K