The volume of a fixed mass of gas at S..T.P is 130cm3, what is the temperature if the new the pressure is 1860mmHg and the volume doubled.
2*130*1860/T = 130*760/273.15
T = 1337 K
or,
2V*2.45P/T' = PV/T
T' = 4.89T = 4.89*273.15 = 1337 K
P1V1/T1 = P2V2/T2
P1 = 760 mm Hg
V1 = 130 cc
T1 = 273 K
P2 = 1860 mm Hg
V2 = 260 cc
T2 = ? K\
Post your work if you get stuck.
To solve this problem, we can use the combined gas law, which relates the initial and final conditions of a gas sample. The formula for the combined gas law 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
T2 = final temperature
We can first convert the units of the given values to ensure they're in a consistent unit system.
Given:
Initial volume (V1) = 130 cm^3
Final volume (V2) = 2 * V1 (since the volume doubled)
Final pressure (P2) = 1860 mmHg
First, we'll convert the initial volume from cm^3 to liters (L):
1 cm^3 = 0.001 L
V1 = 130 cm^3 * 0.001 L/cm^3 = 0.13 L
Next, we'll convert the final pressure from mmHg to atm:
1 atm = 760 mmHg
P2 = 1860 mmHg / 760 mmHg/atm = 2.45 atm
Plugging the given and converted values into the combined gas law, we have:
(P1 * V1) / T1 = (P2 * V2) / T2
Substituting the known values:
(1 atm * 0.13 L) / T1 = (2.45 atm * (2 * 0.13 L)) / T2
Simplifying the equation:
0.13 / T1 = (2.45 * 2 * 0.13) / T2
Tidying up the equation:
0.13 / T1 = 0.637 / T2
To find the temperature T2, we need to cross-multiply and solve for T2:
0.637 * T1 = 0.13 * T2
Dividing both sides by 0.13:
T2 = (0.637 * T1) / 0.13
Now, we can substitute the initial temperature T1 with the value of Standard Temperature and Pressure (STP), which is 273 K.
T2 = (0.637 * 273 K) / 0.13
Calculating T2:
T2 = 1.3333 * 273 K
T2 ≈ 364.665 K
Therefore, the temperature (T2) is approximately 364.665 K when the new pressure is 1860 mmHg and the volume is doubled.