A sample of argon gas has a volume of 735 mL
m
L
at a pressure of 1.20 atm
a
t
m
and a temperature of 112 ∘C
∘
C
. What is the final volume of the gas, in milliliters, when the pressure and temperature of the gas sample are changed to the following, if the amount of gas does not change?
674 mmHg and 274 K
To solve this problem, we can use the combined gas law, which states:
(P1 * V1) / T1 = (P2 * V2) / T2
where P1 is the initial pressure, V1 is the initial volume, T1 is the initial temperature, P2 is the final pressure, V2 is the final volume, and T2 is the final temperature. We are given the initial conditions and asked to find the final volume (V2) under different pressure and temperature conditions.
First, we need to convert the given pressure and temperature to compatible units. Recall that 1 atm = 760 mmHg, so:
1.20 atm * (760 mmHg / 1 atm) = 912 mmHg
Also, we need to convert the temperature from Celsius to Kelvin using the formula K = °C + 273.15.
112 °C + 273.15 = 385.15 K
Now, we plug in the values into the combined gas law equation:
(912 mmHg * 735 mL) / 385.15 K = (674 mmHg * V2) / 274 K
Next, we solve for V2.
V2 = (912 mmHg * 735 mL * 274 K) / (674 mmHg * 385.15 K)
V2 = (229927680 mL K) / (259553.10 mmHg K)
V2 = 885.85 mL
Therefore, the final volume of the gas is approximately 885.85 mL when the pressure and temperature are changed to 674 mmHg and 274 K.