4) Determine the Celsius temperature of 2.49 mol of gas contained in a 2.00-L vessel at a
pressure of 143 kPa.

The temperature of the gas contained in the container is -259.35° C

Explanation:

To find the unit of degrees in Celsius (° C) in this case, the ideal gas law equation is used.

PV = nRT

Where P is the pressure, V is the volume, T is the temperature, n is the number of moles, and R is the universal gas constant.

Solving for T would be: PV / nR

T = ( 1.4113 atm * 2.00 L ) / ( 2.49 mol * 0.08205746 atm * L / mol * K  )

T = 2.8226 / 0.2043

T = 13.8K = -259.35 ° C

The Celsius temperature of the gas is ‪-259.34 ° C

Explanation:

To answer the question, we note that the universal gas equation can be expressed as follows;

P·V = n·R·T

Therefore;

Where:

n = Number of moles  = 2.49 mol

P = Pressure = 143 kPa = 143000 Pa

V = Volume = 2.00 L = ‪0.002 m³

T = Temperature in Kelvin, K = Required

R = Universal Gas Constant =  8.3145 J/(mol·K)

Plugging in the values, we have;

Hence the temperature of the gas = 13.81 K

Converting the Kelvin temperature to Celsius temperature, we have

Celsius temperature = Kelvin temperature - 273.15

Therefore, 13.81 K to Celsius temperature gives;

13.81 - 273.15 = ‪-259.34 ° C.

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