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: Convert pressure to atm

1 atm = 760 torr. So, $P = 3557\ torr\times\frac{1\ atm}{760\ torr}=4.68026\ atm$.

Step2: Identify the ideal - gas constant

The ideal - gas constant $R = 0.0821\ L\cdot atm/(mol\cdot K)$.

Step3: Use the ideal gas law

The ideal gas law is $PV = nRT$. We want to solve for $T$, so $T=\frac{PV}{nR}$. Substitute $P = 4.68026\ atm$, $V = 70.7\ L$, $n = 16.45\ mol$, and $R = 0.0821\ L\cdot atm/(mol\cdot K)$ into the formula. $T=\frac{4.68026\ atm\times70.7\ L}{16.45\ mol\times0.0821\ L\cdot atm/(mol\cdot K)}$. First, calculate the numerator: $4.68026\times70.7 = 331.894382\ L\cdot atm$. Then, calculate the denominator: $16.45\times0.0821=1.350545\ L\cdot atm/K$. Finally, $T=\frac{331.894382\ L\cdot atm}{1.350545\ L\cdot atm/K}=245.74\ K$.

Answer:

$246$ (rounded to three significant figures)