QuestionJune 17, 2025

What is the volume when 524cm3 of a gas at 491 K is cooled to 297 K? 491 K is T_(1) and 524cm3 is V_(1) What variable is V_(2) ? unknown 297K 524cm^3

What is the volume when 524cm3 of a gas at 491 K is cooled to 297 K? 491 K is T_(1) and 524cm3 is V_(1) What variable is V_(2) ? unknown 297K 524cm^3
What is the volume when 524cm3 of a
gas at 491 K is cooled to 297 K?
491 K is T_(1) and 524cm3 is V_(1)
What variable is V_(2) ?
unknown
297K
524cm^3

Solution
4.0(287 votes)

Answer

316.8 \, cm^3 Explanation 1. Identify the formula Use **Charles's Law**: \frac{V_1}{T_1} = \frac{V_2}{T_2}, where V_1 and T_1 are initial volume and temperature, and V_2 and T_2 are final volume and temperature. 2. Substitute known values V_1 = 524 \, cm^3, T_1 = 491 \, K, T_2 = 297 \, K. Substitute these into the formula: \frac{524}{491} = \frac{V_2}{297}. 3. Solve for V_2 Rearrange to find V_2: V_2 = \frac{524 \times 297}{491}. 4. Calculate V_2 Perform the calculation: V_2 \approx 316.8 \, cm^3.

Explanation

1. Identify the formula<br /> Use **Charles's Law**: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$, where $V_1$ and $T_1$ are initial volume and temperature, and $V_2$ and $T_2$ are final volume and temperature.<br /><br />2. Substitute known values<br /> $V_1 = 524 \, cm^3$, $T_1 = 491 \, K$, $T_2 = 297 \, K$. Substitute these into the formula: $\frac{524}{491} = \frac{V_2}{297}$.<br /><br />3. Solve for $V_2$<br /> Rearrange to find $V_2$: $V_2 = \frac{524 \times 297}{491}$.<br /><br />4. Calculate $V_2$<br /> Perform the calculation: $V_2 \approx 316.8 \, cm^3$.
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