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

Answer:

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

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:

$n = \frac{P_{1}V_{1}}{RT_{1}} = \frac{1 atm*62.65 L}{(0.082 L*atm/K*mol)*(0 + 273)K} = 2.80 moles$

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

$T_{2} = \frac{P_{2}V_{2}}{nR} = \frac{612.0 mmHg*\frac{1 atm}{760 mmHg}*78.31 L}{2.80 moles*0.082 L*atm/(K*mol)} = 274.7 K$

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!