Question

A gas has a volume of 62.65L at O degrees Celsius and 1 atm. At what temperature in Celsius would the volume of the gas be 78.31 l at a pressure of 612.0 mmHg

Answer 1

The volume of the gas will be 78.31 L at 1.7 °C.

Step-by-step explanation:

We can find the temperature of the gas by the ideal gas law equation:

`PV = nRT`

Where:

n: is the number of moles

V: is the volume

T: is the temperature

R: is the gas constant = 0.082 L*atm/(K*mol)

From the initial we can find the number of moles:

`n = (P_(1)V_(1))/(RT_(1)) = (1 atm*62.65 L)/((0.082 L*atm/K*mol)*(0 + 273)K) = 2.80 moles`

Now, we can find the temperature with the final conditions:

`T_(2) = (P_(2)V_(2))/(nR) = (612.0 mmHg*(1 atm)/(760 mmHg)*78.31 L)/(2.80 moles*0.082 L*atm/(K*mol)) = 274.7 K`

The temperature in Celsius is:

`T_(2) = 274.7 - 273 = 1.7 ^(\circ) C`

Therefore, the volume of the gas will be 78.31 L at 1.7 °C.

I hope it helps you!