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: Convert pressure to atm

Since $1\ \text{atm} = 760\ \text{torr}$,
$$P = \frac{3557\ \text{torr}}{760\ \text{torr/atm}} \approx 4.680\ \text{atm}$$

Step2: Rearrange ideal gas law

Ideal gas law: $PV = nRT$. Solve for $T$:
$$T = \frac{PV}{nR}$$
where $R = 0.0821\ \frac{\text{L·atm}}{\text{mol·K}}$, $n=16.45\ \text{mol}$, $V=70.7\ \text{L}$

Step3: Substitute values and calculate

$$T = \frac{4.680\ \text{atm} \times 70.7\ \text{L}}{16.45\ \text{mol} \times 0.0821\ \frac{\text{L·atm}}{\text{mol·K}}}$$
$$T = \frac{330.876}{1.350545} \approx 245\ \text{K}$$

Answer:

245 K