A sample of ammonia gas occupies a volume of 1.58 L at 22°C and a pressure of 0.983 atm. What volume will the sample occupy at 1.00 atm and 0°C? (K = oC + 273)
To solve this problem, we can use the combined gas law equation:
(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
Let's plug in the given values:
P1 = 0.983 atm
V1 = 1.58 L
T1 = 22°C + 273 = 295 K
P2 = 1.00 atm
T2 = 0°C + 273 = 273 K
So the equation becomes:
(0.983 atm * 1.58 L) / (295 K) = (1.00 atm * V2) / (273 K)
Now let's solve for V2 (the final volume):
V2 = (0.983 atm * 1.58 L * 273 K) / (295 K * 1.00 atm)
V2 ≈ 1.13 L
Therefore, the volume of the sample of ammonia gas at 1.00 atm and 0°C is approximately 1.13 L.
To solve this problem, we can use the combined gas law equation, which relates the initial and final conditions of pressure, volume, and temperature of a gas sample.
The combined gas law equation is given as:
(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
Now, let's substitute the given values into the equation:
P1 = 0.983 atm
V1 = 1.58 L
T1 = 22°C = 22 + 273 = 295 K
P2 = 1.00 atm
V2 = ?
T2 = 0°C = 0 + 273 = 273 K
Substituting the values into the equation, we get:
(0.983 * 1.58) / (295) = (1.00 * V2) / (273)
Simplifying the equation further, we get:
1.549 / 295 = V2 / 273
Now, cross multiply to solve for V2:
1.549 * 273 = 295 * V2
V2 = (1.549 * 273) / 295
V2 ≈ 1.43 L
So, the sample of ammonia gas will occupy a volume of approximately 1.43 L at 1.00 atm and 0°C.
Use (P1V1/T1)= (P2V2/T2)
T1 and T2 must be in Kelvin. Post your work if you get stuck.