a sample of air has a volume of 140.0 mL a 67 degrees C. at what temperature would its volume be 50.0 mL at constant pressure/

121.43k

so 140.0 x 67/50.0 = 187.6?

No, temperature must be in Kelvin.

Kelvin = 273 + ^o C.

To solve this problem, we can use the combined gas law equation:

(P1 * V1) / T1 = (P2 * V2) / T2

Where:
P1 and P2 are the initial and final pressures (which are constant in this case at constant pressure).
V1 and V2 are the initial and final volumes.
T1 and T2 are the initial and final temperatures.

In this case, we are given:
V1 = 140.0 mL
T1 = 67 degrees C
V2 = 50.0 mL

To find the final temperature (T2), we need to rearrange the equation and solve for T2:

(T2) = (T1 * V2 * P1) / (V1 * P2)

Now we need to convert temperature from degrees Celsius to Kelvin, as the gas law requires the temperatures to be in Kelvin scale. To convert Celsius to Kelvin, we add 273.15 to the Celsius temperature:

T1 in Kelvin = 67 + 273.15 = 340.15 K

Now we can substitute the given values into the equation to solve for T2:

T2 = (340.15 K * 50.0 mL * P1) / (140.0 mL * P2)

Since the pressure is constant, P1 and P2 cancel out. Therefore, we can simplify the equation to:

T2 = (340.15 K * 50.0 mL) / 140.0 mL

T2 = 122.25 K

Therefore, at constant pressure, the temperature at which the volume would be 50.0 mL is 122.25 Kelvin (or -150.9 degrees Celsius).