Answer:
[tex]\boxed{\text{7.50 L}}[/tex]
Explanation:
The temperature and amount of gas are constant, so we can use Boyle’s Law.
[tex]p_{1}V_{1} = p_{2}V_{2}[/tex]
Data:
[tex]\begin{array}{rclrcl}p_{1}& =& \text{3.00 atm}\qquad & V_{1} &= & \text{2.50 L} \\p_{2}& =& \text{1.00 atm}\qquad & V_{2} &= & ?\\\end{array}[/tex]
Calculations:
[tex]\begin{array}{rcl}3.00 \times 2.50 & =& 1.00V_{2}\\7.500 & = & 1.00V_{2}\\V_{2} & = &\textbf{7.50 L}\\\end{array}\\\text{The volume of the gas is } \boxed{\textbf{7.50 L}}[/tex]
Taking into account the Boyle's law, the volume of the gas without a change in temperature will be 7.50 L.
An ideal gas is a theoretical gas that is considered to be composed of point particles that move randomly and do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.
The gas laws are a set of chemical and physical laws that allow determining the behavior of gases in a closed system. The parameters evaluated in these laws are pressure, volume, temperature and moles.
Boyle's law is one of the gas laws, which states that the pressure of a gas in a closed container is inversely proportional to the volume of the container, when the temperature is constant: that is, if the pressure increases, the volume decreases ; whereas if the pressure decreases, the volume increases.
Boyle's law is expressed mathematically as the product of pressure times volume equal to a constant:
P×V= k
Now it is possible to assume that you have a certain volume of gas V1 that is at a pressure P1 at the beginning of the experiment. If you vary the volume of gas to a new value V2, then the pressure will change to P2, and it will be fulfilled:
P1×V1 = P2×V2
In this case, you know:
Replacing in Boyle's law:
3 atm× 2.50 L= 1 atm× V2
Solving:
[tex]V2=\frac{3 atmx2.50 L}{1 atm}[/tex]
V2= 7.50 L
Finally, the volume of the gas without a change in temperature will be 7.50 L.
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