Question

what is the pressure exerted by 5.00 moles of nitrogen gas contained in a 30.0 liter container at 25.0°c? 0.245 atm 3670 atm 4.08 atm 605 atm

Explanation:

Step1: Convert temperature to Kelvin

$$T = 25.0 + 273.15 = 298.15\ \text{K}$$

Step2: Use ideal gas law $PV = nRT$

We know $n = 5.00\ \text{mol}$, $V=30.0\ \text{L}$, $R = 0.0821\ \text{L·atm/(mol·K)}$, $T = 298.15\ \text{K}$. Rearrange for $P$:
$$P=\frac{nRT}{V}$$

Step3: Substitute values

$$P=\frac{5.00\times0.0821\times298.15}{30.0}$$
$$P=\frac{122.4}{30.0}$$
$$P = 4.08\ \text{atm}$$

Answer:

$4.08\ \text{atm}$