What pressure (atm) is needed to confine an ideal gas to 68.0 L after it has expanded from 44.0 L and 2.3 atm at constant temperature?
P1V1=P2V2
solve for P2
P2=P1(V1/V2)=2.3atm(44/68)
Thanks!
To find the pressure (in atm) needed to confine an ideal gas to a certain volume, we can use the combined gas law equation:
(P1 * V1) / T1 = (P2 * V2) / T2
Given:
V1 = 44.0 L (initial volume)
P1 = 2.3 atm (initial pressure)
V2 = 68.0 L (final volume)
T1 = T2 (constant temperature)
Let's substitute the given values into the equation and solve for P2:
(2.3 atm * 44.0 L) / T1 = P2 * 68.0 L / T1
Now, we can cancel out T1:
(2.3 atm * 44.0 L) = P2 * 68.0 L
To isolate P2, divide both sides of the equation by 68.0 L:
P2 = (2.3 atm * 44.0 L) / 68.0 L
Calculating this expression gives us the pressure required to confine the gas to a volume of 68.0 L.