Explanation Write the expression of the mole number of N 2 ( N N 2 ) . N N 2 = m N 2 M N 2 (I) Here, molar mass of N 2 is M N 2 and mass of N 2 is m N 2 Write the expression of the mole number of He ( N He ) . N He = m He M He (II) Here, molar mass of He is M He and mass of He is m He . Write the expression of the mole number of CH 4 ( N CH 4 ) . N CH 4 = m CH 4 M CH 4 (III) Here, molar mass of CH 4 is M CH 4 and mass of CH 4 is m CH 4 . Write the expression of the mole number of C 2 H 6 ( N C 2 H 6 ) . N C 2 H 6 = m C 2 H 6 M C 2 H 6 (IV) Here, molar mass of C 2 H 6 is M C 2 H 6 and mass of C 2 H 6 is m C 2 H 6 . Write the expression of the total number of moles ( N m ) . N m = N N 2 + N He + N CH 4 + N C 2 H 6 (V) Write the expression of the apparent molecular weight of the mixture ( M m ) . M m = m m N m (VI) Write the expression of the constant pressure specific heat of the mixture ( c p ) . c p = mf N 2 c p , N 2 + mf He c p , H e + mf C H 4 c p , C H 4 + mf C 2 H 6 c p , C 2 H 6 (VII) Here, mole fraction of N 2 ,He,CH 4 , and C 2 H 6 is mf N 2 , mf He , mf CH 4 , and mf C 2 H 6 , specific heat of N 2 is c p , N 2 , specific heat of He is c p , H e , specific heat of CH 4 is c p , C H 4 , and specific heat of C 2 H 6 is c p , C 2 H 6 . Write the expression of the apparent gas constant of the mixture ( R ) . R = R u M m (VIII) Here, universal gas constant is R u . Write the expression of the constant volume specific heat ( c v ) . c v = c p − R (IX) Write the expression of the mass of the container ( m m ) . m m = P 1 V m R T 1 (X) Here, volume of the mixture is V m , initial pressure and temperature is P 1 and T 1 . Use energy balance on the system to calculate final temperature ( T 2 ) . W sh,in = m m c v ( T 2 − T 1 ) T 2 = T 1 + W sh,in m m c v (XI) Here, amount of work done on the mixture is W sh,in . Write the expression of the final pressure from an ideal gas. P 2 = P 1 T 2 T 1 (XII) Conclusion: Refer Table A-1, “Molar mass, gas constant, and critical point properties”, obtain the molar masses of N 2 , He , CH 4 and C 2 H 6 as 28.0 kg / kmol , 4.0 kg / kmol , 16.0 kg / kmol and 30.0 kg / kmol . Substitute 15 kg for m N 2 and 28.0 kg / mol for M N 2 in Equation (I). N N 2 = 15 kg 28.0 kg / mol = 0.5357 kmol Substitute 5 kg for m He and 4.0 kg / mol for M He in Equation (II). N He = 5 kg 4.0 kg / mol = 1.25 kmol Substitute 60 kg for m CH 4 and 16.0 kg / mol for M CH 4 in Equation (III). N CH 4 = 60 kg 16

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A substance with fixed chemical composition throughout is called a pure substance like water, air, and nitrogen. A pure substance does not have to be a single element or compound. A mixture of two or more phases of a pure substance is also a pure substan…

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Chapter 13.3, Problem 56P

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The mixture’s final temperature and pressure.

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