Sodium azide is used in automobile air bags. Assuming that crashes would happen

Question

Sodium azide is used in automobile air bags. Assuming that crashes would happen at STP, what amount of NaN_3 is needed to inflate the air bag 56 L during a crash? The overall reaction for NaN_3 in air bags is: NaN_3(s) rightarrow Na_(s) + N_2(g) b) A 1.00 L sample of gas at 100 degree C and 0.789 atm contains 50% helium and 50% xenon by mass. What are the partial pressures of the individual gases? c) Calculate the Delta E for the following processes: 1) q = 100 J and w = 1 kJ 2) The system absorbs 530 Joules of heat and does 150 Joules of work on the surroundings?

Explanation / Answer

(a) at STP, T=0OC =273 K

                p = 1 atm

               V= 56 litre

               R= 0.082057 L atm mol-1 K-1

By ideal gas law we have

PV=NRT      ....................................................(1)

Substituting the known values in the above equation

      1 * 56=N*0.082057*273

              N=2.499 moles

             Molecular weight of NaN3 =23+2*14=51

                  Moles=Weight/Molecuar weight

                 Amount of NaN3 required =2.4998*51=127.49 grams

(b) T=100OC =373 K

                p = 0.789atm

               V= 1 litre

               R= 0.082057 L atm mol-1 K-1

            PV =NRT

Substituting the known values in the above equation we get,

1*0.789 =N*0.082057*373

             N=0.0257 moles

the above moles are total moles ,i.e, sum of moles of xenon and helium

       Nxe +NHe =0.0257            ..................(2)

      (W1/Mwxe)   +(W2/MwHe)=0.0257

      (W1/131)   +(W2/4)=0.0257

but 50 % mass of xenon and helium is added .Therefore W1=W2 =W

   W(1/131+1/4)=0.0257

W=0.09975,

NXe =0.09975/131=7.6145*10-4

NHe=0.09975/4=0.0249375

Lets us calculate now the mole fraction of each component

Mole fraction of xenon =Yxe=7.6145*10-4/(7.6145*10-4+0.0249375)=0.0296

Mole fraction of helium =YHe=0.0249375/(0.0249375+7.6145*10-4)=0.9704

For idea gas law Partial pressure =Yi*P

where Yi=mole fraction of species 'i'

and P=Total pressure of the system

Partial pressure of Xenon =Yxe*P =0.0296*0.789=0.02335 atm

Partial pressure of Xenon =YHe*P=0.9704*0.789=0.7656 atm

(c) E =Q-W

            Q = postive when heat given to the system

               = negative when heat given to surrounding

           W= postive when works is done by the system

             = negative when work in done on the system

1.)q=100 J W=1 kJ=1000 J

     E= 100-1000= -900 J

2.) system absorbs heat .i.e. heat is given to the system , q=+530J

          150 joules of work is done on the surrounding ,i.e, work is done by the system on the system = +150 Joules

     E=530-150=380 Joules

       

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