Question

part a
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kcv simple gas laws and ideal gas law; read
section 6.3. you can click on the review link to
access the section in your etext.
what will its volume be if an additional 0.152 mol of gas is added to the cylinder?
(assume constant temperature and pressure.)
a cylinder with a moveable piston contains 0.469
mol of gas and has a volume of 171 ml.
express your answer in milliliters to three significant figures.
view available hint(s)
v₂ = ml

Explanation:

Step1: Find the new amount of gas

The initial amount of gas \(n_1 = 0.469\space mol\). The additional amount of gas is \(0.152\space mol\). So the new amount of gas \(n_2=n_1 + 0.152\space mol=0.469+ 0.152=0.621\space mol\)

Step2: Apply Avogadro's law

Avogadro's law is \(V_1/n_1 = V_2/n_2\), where \(V_1 = 171\space mL\), \(n_1 = 0.469\space mol\), \(n_2=0.621\space mol\). We can solve for \(V_2\):

$$V_2=\frac{V_1\times n_2}{n_1}$$

Substitute the values:

$$V_2=\frac{171\times0.621}{0.469}$$

$$V_2=\frac{106.191}{0.469}$$

$$V_2 = 226.42\space mL\approx226\space mL$$

Answer:

\(226\space mL\)