Question

part a
what is the final temperature of a 2.50 l system when its volume is reduced to 1.60 l if the initial temperature was 298 k? (assume the pressure is constant and no gas particles are lost)
107 k
191 k
477 k
466 k
none of the above
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Explanation:

Step1: Identify the gas law

Since pressure is constant, we use Charles's Law, which states that $\frac{V_1}{T_1} = \frac{V_2}{T_2}$, 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: List the given values

We have $V_1 = 2.50\ L$, $T_1 = 298\ K$, and $V_2 = 1.60\ L$. We need to find $T_2$.

Step3: Rearrange the formula to solve for $T_2$

From $\frac{V_1}{T_1} = \frac{V_2}{T_2}$, we can rearrange to get $T_2 = \frac{V_2 \times T_1}{V_1}$.

Step4: Substitute the values into the formula

Substitute $V_1 = 2.50\ L$, $T_1 = 298\ K$, and $V_2 = 1.60\ L$ into the formula: $T_2 = \frac{1.60\ L \times 298\ K}{2.50\ L}$.

Step5: Calculate the result

First, calculate the numerator: $1.60 \times 298 = 476.8$. Then divide by the denominator: $\frac{476.8}{2.50} = 190.72\ K \approx 191\ K$.

Answer:

191 K (corresponding to the option "191 K")