Question

if 22.5 mol of an ideal gas is at 6.53 atm at 97.50 °c, what is the volume of the gas? v =

Explanation:

Step1: Convert temperature to Kelvin

$$T=(97.50 + 273.15)\text{ K}=370.65\text{ K}$$

Step2: Use ideal gas law \(PV = nRT\) to find volume \(V\)

The ideal gas constant \(R = 0.0821\text{ L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}\), \(n = 22.5\text{ mol}\), \(P=6.53\text{ atm}\)
From \(V=\frac{nRT}{P}\), substitute the values:
$$V=\frac{22.5\text{ mol}\times0.0821\text{ L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}\times370.65\text{ K}}{6.53\text{ atm}}$$
First calculate the numerator: \(22.5\times0.0821\times370.65=22.5\times30.420365 = 684.4582125\)
Then divide by \(P\): \(V=\frac{684.4582125}{6.53}\approx105\text{ L}\)

Answer:

\(105\text{ L}\)