The volume of an ideal gas is adiabatically reduced from 200Lto74.3L. The initial pressure and temperature are 1.00atmand 300K.The final pressure is 4.00atm.

  1. Is the gas monoatomic, diatomic or polyatomic?
  2. What is the final temperature?
  3. How many moles are in the gas?

Short Answer

Expert verified
  1. The gas is diatomic.
  2. The final temperature is Tf=446K.
  3. The number of moles in the gas is.n=8.10×103moles.

Step by step solution

01

Given

  • Initial volume of the gas is Vi=200L
  • Final volume of the gas is Vf=74.3L
  • Initial pressure ispi=1.00atm
  • Final pressure ispf=4.00atm

Initial temperature is TI=300K

02

Understanding the concept

In case of adiabatic process,

pVγ=Constant

Here p is the pressure, V is the volume and γis the ratio specific heat capacity at constant pressure to the specific heat capacity at constant volume.

γ=CPCV

Here Cpis specific heat capacity at constant pressure and CVis specific heat capacity at constant volume.

03

(a) Find out if the gas is monoatomic, diatomic or polyatomic

For adiabatic process,

piViγ=pfVfγ

Hence,

pfpi=ViVfγ

Substitute 4.00atm for pf ,1.00atm for pi, 200Lfor VIand 74.3L for Vf into the above equation,

4.001.00=200743γ4=(2.691)y

Taking log on both sides we get,

In(4)=γIn(2.691)γ=In(4)In(2.691)=1.4=75

From the table 19-3 we can conclude that it is a diatomic gas.

04

(b) Calculate the final temperature

Ideal gas law equations for initial and final states are

piVi=nRTi…… (1)

pfVf=nRTf…… (2)

Dividing equation (2) by (1) we get,

localid="1663961348430" pfVfpiVi=nRTfnRTi

pfVfpiVi=TfTi

Tf=TipfVfpiVi

Substitute 300K for Ti, 4atm for pf, 74.3L for Vf , 1atm for pi and 200L for Vi into the above equation,

localid="1662532760854" TT=300(4×74.3)1×200=445.8=446K

Therefore the final temperature is 446K .

05

(c) Calculate the no. of moles in the gas

Using ideal gas law we get,

piVi=nRTin=(piVi)/(RTi)

Substitute1.01×105pa for pi, 200L for Vi,8.3145J/molK for R and 300K for TIinto the above equation,

n=1.01×105×2008.314×300=8.10×103moles

The number of moles in the gas is n= 8.10×103 moles.

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