If 3.86 mol of an ideal gas has a pressure of 2.10 atm and a volume of 77.27 L, what is the temperature of the sample in degrees Celsius?

Answer 1

Ideal Gas Law

PV = nRT

P = pressure = 2.10 atm

V = volume = 77.27 L

n = moles = 3.86

R = gas constant = 0.0821 Latm/Kmol

T = temperature in Kelvin = ?

Solving for T:

T = PV/nR = (2.10)(77.27)/(3.86)(0.0821)

T = 512.04 K

To convert to ºC, subtract 273.15K

T = 512.04 - 273.15 = 238.89ºC

Answer 2

The ideal gas law is: PV = NRT.

In your case, P = 2.10 atm, N = 3.86 mol, V = 77.27 L, R (which is a constant) = 0.08205, and T = an unknown number of degrees kelvin.

Solving (2.10)(77.27) = (3.86)(0.08205)(T) evaluates to T = 512.3471. This is in Kelvin however, and to convert Kelvin to celsius, we must subtract 273.15 from the Kelvin temperature. 512.347-273.15 = 239.197 degrees Celsius.

Hope this helps!