(BOYLE'S LAW)
a gas occupies 5 L at 6.00 atm. what will be the initial volume of this gas if the pressure becomes 3.00 atm?
Share
(BOYLE'S LAW)
a gas occupies 5 L at 6.00 atm. what will be the initial volume of this gas if the pressure becomes 3.00 atm?
Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
Answer:
The volume of this gas : 0.52 L
Further explanation
Some of the laws regarding gas, can apply to ideal gas (volume expansion does not occur when the gas is heated), among others
Boyle's law at constant T, P = 1 / V
Charles's law, at constant P, V = T
Avogadro's law, at constant P and T, V = n
So that the three laws can be combined into a single gas equation, the ideal gas equation
\rm PV=nRTPV=nRT
In Boyle's law, it is stated that
In a closed system, the absolute pressure by an ideal gas is inversely proportional to the volume it occupies if the temperature is constant
So it can be stated
\rm \large{\boxed{\bold{P_1.V_1=P_2.V_2}}
A gas occupies 1.56 L at 1.00 atm. and the pressure becomes 3.00 atm
So :
Initial conditions:
P₁ = 1 atm
V₁ = 1.56 L
Final conditions
P₂ = 3 atm
The volume of this gas (V₂) :
\begin{gathered}\rm V_2=\dfrac{P_1.V_1}{P_2}\\\\V_2=\dfrac{1\times 1.56}{3}\\\\V_2=\boxed{\bold{0.52~L}}\end{gathered}
V
2
=
P
2
P
1
.V
1
V
2
=
3
1×1.56
V
2
=
0.52 L
Answer:
PV = constant
(1.56 L) (1.00 atm) = (3.00 atm) (x); 0.520 L