1) If a 10. m3 volume of air (acting as an ideal gas) is at a pressure of 760 mm and a temperature of 27 degrees Celsius is taken to a high altitude where the pressure is 400. mm Hg and a temperature of -23 degrees Celsius, what volume will it occupy? (Hint: remember the temperature must be in Kelvin)
Would you use the same formula for this one? V2 = V1 (P1/P2)(T2/T1)
Yes, use the same formula but be sure you convert the temperatures to K.
T1 = 300 K and T2 = 250 K.
Yes, you can use the formula V2 = V1 (P1/P2)(T2/T1) to solve this problem. The formula is known as the ideal gas law, which states that the product of pressure and volume is directly proportional to the product of temperature and amount of gas, provided the gas behaves ideally.
To solve the problem, we can plug in the given values into the formula and solve for V2, which represents the final volume.
Given:
V1 = 10 m3
P1 = 760 mm Hg
P2 = 400 mm Hg
T1 = 27 degrees Celsius
T2 = -23 degrees Celsius
However, we need to convert the temperatures to Kelvin since the ideal gas law requires temperatures to be in Kelvin. The Kelvin temperature scale is obtained by adding 273.15 to the Celsius temperature.
Converting the temperatures:
T1 = 27 + 273.15 = 300.15 K
T2 = -23 + 273.15 = 250.15 K
Now, we can substitute the values into the formula:
V2 = V1 (P1/P2)(T2/T1)
= 10 m3 (760 mm Hg / 400 mm Hg) (250.15 K / 300.15 K)
Simplifying the equation:
V2 = 10 m3 * 1.9 * 0.8339
= 15.8385 m3
Therefore, the volume of air at the high altitude would occupy approximately 15.8385 m3.