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?

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

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :