249 mol of argon gas is admitted to an evacuated 18847 cm3 c

2.49 mol of argon gas is admitted to an evacuated 188.47 cm3 container at 38.32oC. The gas then undergoes an isochoric heating to a temperature of 226.78oC. What is the final pressure?

Solution

Number of moles n = 2.49 mol

Initial Volume V= 188.47 cm³ = 188.47x10 ^-6 m ³

Initial Temprature T = 38.32^oC = 38.32 + 273 = 311.32 K

Initial pressure P = nRT / V

                          = (2.49)(8.314)(311.32 ) /(188.47 x10 ^-6 )

                          = 34.195 x10 ⁶ Pa

Final temperature T \' = 226.78^oC = 226.78 +273 = 499.78 K

The final pressure P \' = ?

In isochoric process,i.e., at constant volume,

P \' /P = T \' / T

         = 499.78 / 311.32

         = 1.605

     P \' = 1.605 P

          = 1.605 x 34.195 x10 ⁶ Pa

          = 54.89 x10 ⁶ Pa