An ideal gas having a volume of 10 Times 103 m3 at 40 degree

An ideal gas having a volume of 1.0 Times 10^-3 m^3 at, 40 degree C and a pressure of 1 atm expands until its volume is 1.5 Times 10^-3 m^3 and its pressure is 1.053 atm. Noting that PV = nRT, T(K) = T(degree C) + 273, and that 1 atm = 1.013 Times 10^5 N/m^2, please find: The number of moles of the gas. The final temperature of the gas in degree C.

Solution

Volume V = 10 -3 m 3

Initial temprature T = 40 o C = 40 + 273 = 313 K

Initial pressure P = 1 atm = 1.01 x10 5 Pa

Number of moles n = ?

from the relation PV = nRT

number of moles n = PV/RT

                           =(1.01 x10 5)(10 -3) /(8.314 x313)

                           = 0.038812 mol

(B). Final volume V \' = 1.5 x10 -3 m 3

Final pressure P \' = 1.053 atm

Final temprature T \' = ?

We know PV/T = P \' V \' / T \'

From this T \' = P \' V \' T / PV

                    =(1.053 atm)(1.5 x10 -3 )(313) /(1 atm)(10 -3 )

                    = 494.38 K

                   = (494.38 -273 ) o C

                   = 221.38 o C

 An ideal gas having a volume of 1.0 Times 10^-3 m^3 at, 40 degree C and a pressure of 1 atm expands until its volume is 1.5 Times 10^-3 m^3 and its pressure is

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