A sample of a gas occupies a volume of 1158 mL at 59ºC and 0.76 atm. What would be the new volume if temperature decreases to 40ºC and pressure increases to 1.40 atm?
use
P1V1/T1=P2V2/T2
To find the new volume, we can use the combined gas law equation, which combines Boyle's Law, Charles' Law, and Gay-Lussac's Law. The equation is as follows:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where:
P1 = initial pressure
V1 = initial volume
T1 = initial temperature
P2 = final pressure
V2 = final volume (which we need to find)
T2 = final temperature
Let's plug in the given values:
P1 = 0.76 atm
V1 = 1158 mL
T1 = 59ºC -> convert to Kelvin by adding 273.15 (59 + 273.15 = 332.15 K)
P2 = 1.40 atm
T2 = 40ºC -> convert to Kelvin (40 + 273.15 = 313.15 K)
Now we can rearrange the equation to solve for V2:
V2 = (P1 * V1 * T2) / (P2 * T1)
Substituting the given values into the equation:
V2 = (0.76 atm * 1158 mL * 313.15 K) / (1.40 atm * 332.15 K)
Before calculating, let's convert mL to liters by dividing by 1000:
V2 = (0.76 atm * 1.158 L * 313.15 K) / (1.40 atm * 332.15 K)
Now we can cancel out the units:
V2 = 0.680 L
Therefore, the new volume would be approximately 0.680 L when the temperature decreases to 40ºC and the pressure increases to 1.40 atm.