Question

a 7.50 liter sealed jar at 18°c contains 0.25 moles of gas. what is the pressure in the container?

Explanation:

Step1: Convert temperature to Kelvin

$$T = 18 + 273.15 = 291.15\ K$$

Step2: Use ideal gas law \(PV = nRT\)

We need to find \(P\), and we know \(V=7.50\ L\), \(n = 0.25\ mol\), \(R=0.0821\ L\cdot atm/(mol\cdot K)\), \(T = 291.15\ K\).
Rearrange the formula for \(P\): \(P=\frac{nRT}{V}\)
Substitute the values:
$$P=\frac{0.25\ mol\times0.0821\ L\cdot atm/(mol\cdot K)\times291.15\ K}{7.50\ L}$$

Step3: Calculate the value

First calculate the numerator: \(0.25\times0.0821\times291.15=0.25\times23.903415 = 5.97585375\)
Then divide by \(7.50\): \(P=\frac{5.97585375}{7.50}=0.7967805\approx0.796\ atm\)

Answer:

\(0.796\ atm\)