Question

According to Boyle's law, the relation = p1. V2 ——=— P2 V1holds for pressures p1 and p2 and volumes V1, and V2 of a gas at constant temperature. Find V1, if p₁ = 36.5 kPa,P2=84.1 kpa, and v2=.0447 m^3(Simplify your answer. Type an integer or decimal rounded to four decimal places as needed.)

Answer 1

SOLUTION

From the formula given we have that

$\begin{gathered} \frac{P_1}{P_2}=\frac{V_2}{V_1} \\ Now\text{ substitute } \\ P_1=36.5,P_2=84.1,V_2=0.0447\text{ into the equation, we have } \end{gathered}$

V1 as

$\begin{gathered} \frac{36.5}{84.1}=\frac{0.0447}{V_1} \\ cross\text{ multiplying we have } \\ 36.5V_1=0.0447\times84.1 \\ V_1=\frac{0.0447\times84.1}{36.5} \\ V_1=0.1029937 \end{gathered}$

Hence the answer is

$V_1=0.1030\text{ }m^3$