1.45 moles of helium gas occupies a volume 18.5 L at a pressure of .85 atm. What is the temperature of the gas in K and °C?

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The temperature of the gas is 132.25K and -140.75°C

Explanation:

Given that:

Volume of helium gas (V) = 18.5L

Temperature of helium gas (T) = ?

Pressure (P) = 0.85 atm

Number of moles (n) = 1.45 moles

Molar gas constant (R) is a constant with a value of 0.082 atm L K-1 mol-1

Then, apply the formula for ideal gas equation

pV = nRT

0.85atm x 18.5L = 1.45 moles x 0.082 atm L K-1 mol-1 x T

15.725 atm•L = 0.1189 atm•L•K-1 x T

T = 15.725 atm•L / 0.1189 atm•L•K-1

T = 132.25K

Given that temperature in Kelvin is 132.25, convert it to Celsius by subtracting 273

So, T(in Celsius) = 132.25K - 273

= -140.75°C

Thus, the temperature of the gas is 132.25K and -140.75°C

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