a 6 l container of hydrogen gas, originally at 298 k and 749 torr, is heated until a volume of 8 l and a…

a 6 l container of hydrogen gas, originally at 298 k and 749 torr, is heated until a volume of 8 l and a pressure of 1150 torr is reached. what is the final kelvin temperature?

a 6 l container of hydrogen gas, originally at 298 k and 749 torr, is heated until a volume of 8 l and a pressure of 1150 torr is reached. what is the final kelvin temperature?

Answer

Explanation:

Step1: Recall the combined - gas law

The combined - gas law is $\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$, where $P_1$, $V_1$, $T_1$ are the initial pressure, volume and temperature respectively, and $P_2$, $V_2$, $T_2$ are the final pressure, volume and temperature respectively.

Step2: Identify the given values

$P_1 = 749$ Torr, $V_1=6$ L, $T_1 = 298$ K, $P_2 = 1150$ Torr, $V_2 = 8$ L. We need to find $T_2$.

Step3: Rearrange the combined - gas law for $T_2$

$T_2=\frac{P_2V_2T_1}{P_1V_1}$.

Step4: Substitute the values into the formula

$T_2=\frac{1150\times8\times298}{749\times6}$. First, calculate the numerator: $1150\times8\times298 = 1150\times2384=2741600$. Then, calculate the denominator: $749\times6 = 4494$. Now, divide: $T_2=\frac{2741600}{4494}\approx610$ K.

Answer:

610 K