A sample of argon gas occupies a volume of 950 mL at 25.0°C. What
volume will the gas occupy at 50.0°C if the pressure remains constant?
(V1/T1)=(V2/T2)
Don't forget to use Kelvin for T1 and T2.
(0.95 L x 323 K) / 298 K = 1.03 L of Argon gas
To find the volume of the argon gas at 50.0°C, we can use the combined gas law equation:
P1V1 / T1 = P2V2 / T2
Where:
P1 and P2 are the initial and final pressures (in this case, the pressure remains constant),
V1 and V2 are the initial and final volumes,
T1 and T2 are the initial and final temperatures.
We are given:
V1 = 950 mL
T1 = 25.0°C
T2 = 50.0°C
Now, we need to convert the temperatures from Celsius to Kelvin since the temperature must be in Kelvin for the equation:
T1(K) = T1(°C) + 273.15
T2(K) = T2(°C) + 273.15
So, T1(K) = 25.0 + 273.15 = 298.15 K
T2(K) = 50.0 + 273.15 = 323.15 K
Now we can plug in the values into the combined gas law equation and solve for V2:
P1V1 / T1 = P2V2 / T2
Since the pressure remains constant, we can simplify the equation:
V1 / T1 = V2 / T2
V2 = (V1 * T2) / T1
Now we substitute the given values:
V2 = (950 mL * 323.15 K) / 298.15 K
Calculating this, we find:
V2 = 1031.74 mL
Therefore, the volume of the argon gas at 50.0°C is approximately 1031.74 mL, considering a constant pressure.