---Calculate the average molecular kinetic energy of a 0.250 molequantity of an

Question

---Calculate the average molecular kinetic energy of a 0.250 molequantity of an ideal gas at a temperature of -8.00oC.

---A 0.400 mole sample of an ideal gas is at a temperature of 297 Kat atmospheric pressure. Calculate the volume occupied by thegas.

---A sample of an ideal gas is both compressed and cooled. Thegiven variables for the gas are: P1 = 90.0 kPa,V1 = 0.0800 m3,V2 = 0.0400 m3,T1 = 300. K, T2 = 260. K.Calculate the final pressure.

Explanation / Answer

1) temperature = -8.00o C = 265 K KE = (3/2)kT where k = boltzman constatnt = 1.38*10-23 J/K => KE = (1.5)*( 1.38*10-23 )(265) =548.55*10-23 J => KE = 548.55*10-23 J =5.4855*10-21 J 2)given, n = 4 mol T = 297K P = 1atm V=? we know that, By Ideal gas equation PV =nRT          where R = 0.0821LatmK1mol1 => 1*V = 4*0.0821*297 => V = 97.5348 L 3)Given : P1 = 90.0 kPa, V1= 0.0800 m3, V2 = 0.0400m3, T1 = 300. K,T2 = 260. K Ideal gas equation at initial stage P1V1 =nRT1 ---------eq1 Ideal GAs equatiuon at final stage P2V2 =nRT2   ------eq 2 dividing eq1 by eq2 , we get P1V1 /P2V2 =T1/T2 putting all values [90*0.08]/[P2 *0.04] = 300/260 => P2 = 90*2*260/300 = 156kPa

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