If 3.11 mol of an ideal gas has a pressure of 2.91 atm and a volume of 78.13 L, what is the temperature of the sample in degrees Celsius?

Answer: 617.35 °C

Explanation:

Using the ideal gas law,

PV = nRT

P = Pressure in (atm or pascal)

V = Volume in (litres or cubic metres)

n = number of moles.

R = Gas constant in (0.0821L. atm/mol. K or 8.314J/mol/K)

T = Temperature in degree Kelvin or Celsius.

PV = nRT

P = 2.91atm, V = 78.13L, n = 3.11

R = 0.0821L. atm/mol. K

T = ?

T = 2.91 x 78.13 / 3.11 x 0.0821 = 227.3583 / 0.255331 = 890.5 K

Converting from Kelvin to Celsius, using: °C = K - 273.15

= 890.5 - 273.15

°C = 617.35