Question

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

Explanation:

Step1: Convert temperature to Kelvin

The ideal gas law uses absolute temperature. We convert \(T = 25.0^{\circ}C\) to Kelvin. The formula is \(T(K)=T(^{\circ}C)+ 273.15\). So, \(T=(25.0 + 273.15)K=298.15K\)

Step2: Recall the ideal gas law

The ideal gas law is \(PV = nRT\), where \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the ideal gas constant (\(R = 0.0821\frac{L\cdot atm}{mol\cdot K}\)), and \(T\) is temperature. We need to solve for \(P\), so \(P=\frac{nRT}{V}\)

Step3: Substitute the values

Given \(n = 5.00mol\), \(V=30.0L\), \(R = 0.0821\frac{L\cdot atm}{mol\cdot K}\), and \(T = 298.15K\). Substitute into the formula:

$$P=\frac{5.00mol\times0.0821\frac{L\cdot atm}{mol\cdot K}\times298.15K}{30.0L}$$

First, calculate the numerator: \(5.00\times0.0821\times298.15=(5.00\times0.0821)\times298.15 = 0.4105\times298.15\approx122.4\)
Then, divide by \(V = 30.0L\): \(P=\frac{122.4}{30.0}=4.08atm\)

Answer:

\(4.08atm\)