Question

what is the temperature (in k) of 16.45 moles of methane gas in a 70.7 l container at 3,557 torr? ? k

Explanation:

Step1: Recall ideal gas law

The ideal gas law is $PV = nRT$, where $P$ is pressure, $V$ is volume, $n$ is moles, $R$ is gas constant, $T$ is temperature. Rearrange to solve for $T$: $T = \frac{PV}{nR}$.

Step2: Convert pressure to atm

1 atm = 760 torr, so $P = \frac{3557}{760} \approx 4.680$ atm.

Step3: Choose R value

Use $R = 0.0821 \frac{L \cdot atm}{mol \cdot K}$.

Step4: Calculate T

Substitute values: $T = \frac{(4.680)(70.7)}{(16.45)(0.0821)}$.

First compute numerator: $4.680 \times 70.7 \approx 330.976$.
Denominator: $16.45 \times 0.0821 \approx 1.351$.
$T \approx \frac{330.976}{1.351} \approx 245$.

Answer:

245 K