QuestionJune 5, 2025

1. The pressure of a 6.02 mol sample of F_(2) in a 90.729 L container is measured to be 7.06 atm. What is the temperature of this gas in kelvins?

1. The pressure of a 6.02 mol sample of F_(2) in a 90.729 L container is measured to be 7.06 atm. What is the temperature of this gas in kelvins?
1. The pressure of a 6.02 mol sample of F_(2) in a 90.729 L container is measured to be 7.06 atm.
What is the temperature of this gas in kelvins?

Solution
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Answer

1297.57 K Explanation 1. Use Ideal Gas Law The ideal gas law is given by **PV = nRT**. Here, P = 7.06 \text{ atm}, V = 90.729 \text{ L}, n = 6.02 \text{ mol}, and R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1}. 2. Solve for Temperature T Rearrange the formula to solve for T: T = \frac{PV}{nR}. Substitute the values: T = \frac{(7.06)(90.729)}{(6.02)(0.0821)}. 3. Calculate the Temperature Perform the calculation: T = \frac{640.94754}{0.494042} \approx 1297.57 \text{ K}.

Explanation

1. Use Ideal Gas Law<br /> The ideal gas law is given by **$PV = nRT$**. Here, $P = 7.06 \text{ atm}$, $V = 90.729 \text{ L}$, $n = 6.02 \text{ mol}$, and $R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1}$.<br /><br />2. Solve for Temperature $T$<br /> Rearrange the formula to solve for $T$: $T = \frac{PV}{nR}$. Substitute the values: $T = \frac{(7.06)(90.729)}{(6.02)(0.0821)}$.<br /><br />3. Calculate the Temperature<br /> Perform the calculation: $T = \frac{640.94754}{0.494042} \approx 1297.57 \text{ K}$.
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