Assuming the temperature is kept constant during this change, you can use Boyle's Law to answer the question. The law relates the products of volume (V) and pressure (P) before and after:
[tex]P_1V_1=P_2V_2[/tex]
Given V1, P1, and V2:
[tex]P_2 = \frac{P_1V_1}{V_2} = \frac{1.75atm\cdot 1.25l}{3.15l}=0.69atm[/tex]
So, the pressure after the volume change, while the temperature is constant, will be 0.69 atmospheres. This is intuitive: the volume has increased with same amount of ideal gas, and so the pressure went down.
The general gas equation, commonly known as the ideal gas law, is a state equation for a hypothetical ideal gas. The pressure is if the volume is changed to 3.15 L will be 0.69 atm.
The general gas equation, commonly known as the ideal gas law, is a state equation for a hypothetical ideal gas.
Despite its flaws, the ideal gas law provides a decent approximation of the behavior of various gases under a variety of situations.
If the ideal gas equation is applied and equated for the two gases having temperature constant we will obtain the equation as;
p₁v₁=p₂v₂
[tex]\rm p_2= \frac{p_1v_1}{v_2} \\\\ \rm p_2= \frac{1.75 \times 1.25}{3.15} \\\\ \rm p_2= 0.69 atm[/tex]
Hence the pressure is if the volume is changed to 3.15 L will be 0.69 atm.
To learn more about the ideal gas equation refer to the link;
https://brainly.com/question/4147359
