A sample of argon gas has a volume of 715 mL at a pressure of 1.34 atm and a temperature of 144 Celsius.
What is the volume of the gas in milliliters when the pressure and temperature of the gas sample are changed to 622 mmHg and 321 K?
(P1V1)/T1 = (P2V2)/T2
Use Kelvin for both T1 and T2. You will need to change the pressure to the same units--it makes no difference which one you change.
902 mL
I obtained 901 mL
To find the volume of the gas when the pressure and temperature are changed, we can use the combined gas law equation:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Given:
P1 = 1.34 atm (initial pressure)
V1 = 715 mL (initial volume)
T1 = 144 °C + 273.15 = 417.15 K (initial temperature)
P2 = 622 mmHg (final pressure)
T2 = 321 K (final temperature)
First, let's convert the given pressure from mmHg to atm:
1 mmHg = 0.00131579 atm (approximately)
P2 = 622 mmHg * 0.00131579 atm/mmHg ≈ 0.8187 atm
Now we can substitute the given values into the combined gas law equation:
(1.34 atm * 715 mL) / (417.15 K) = (0.8187 atm * V2) / (321 K)
Next, we rearrange the equation to solve for V2, the final volume:
(1.34 atm * 715 mL * 321 K) / (0.8187 atm * 417.15 K) = V2
Simplifying the equation gives us:
V2 ≈ 625 mL
Therefore, the volume of the gas sample will be approximately 625 mL when the pressure and temperature are changed to 622 mmHg and 321 K.