At 650 degree C and a pressure of 690mm HG, the 0.927g sample has a volume of 194cm3.
How do I calculate the volume at the same temp but at a pressure of 760mm Hg?
Also, how do I calculate the volume at the same pressure but at a temp of 0 degree C?
Thanks
if it is a gas, you can use the combined gas law.
P1V1/T1=P2v2/T2
change temsp to K.
To calculate the volume at the same temperature but at a pressure of 760 mmHg, we can use the combined gas law equation:
P1V1/T1 = P2V2/T2
You are given:
P1 = 690 mmHg (initial pressure)
V1 = 194 cm³ (initial volume)
T1 = 650 °C (initial temperature, which needs to be converted to Kelvin)
To convert temperature from °C to Kelvin, you can use the formula:
T(K) = T(°C) + 273.15
So, for the initial temperature:
T1(K) = 650 °C + 273.15 = 923.15 K
The given final pressure is:
P2 = 760 mmHg
Now, we can rearrange the equation and plug in the values:
(690 mmHg)(194 cm³)/(923.15 K) = (760 mmHg)(V2)/(923.15 K)
Simplifying the equation, we get:
V2 = [(690 mmHg)(194 cm³)/923.15 K] * (923.15 K/760 mmHg)
V2 ≈ 201.83 cm³
So, the volume at the same temperature but at a pressure of 760 mmHg is approximately 201.83 cm³.
To calculate the volume at the same pressure but at a temperature of 0 °C, we can use the same combined gas law equation:
P1V1/T1 = P2V2/T2
The given initial pressure is:
P1 = 690 mmHg
The given initial volume is:
V1 = 194 cm³
The given final temperature is:
T2 = 0 °C (which needs to be converted to Kelvin)
Using the conversion formula:
T2(K) = T2(°C) + 273.15
T2(K) = 0 °C + 273.15 = 273.15 K
Now, we can rearrange the equation and plug in the values:
(690 mmHg)(194 cm³)/(923.15 K) = P2(194 cm³)/(273.15 K)
Simplifying the equation, we get:
P2 = [(690 mmHg)(194 cm³)/923.15 K] * (273.15 K/194 cm³)
P2 ≈ 385.71 mmHg
So, the volume at the same pressure but at a temperature of 0 °C is approximately 385.71 mmHg.