2 mole of an ideal monoatomic gas is mixed with 3 mole of an ideal diatomic gas at room temperature then γ for the mixture is

Question

2 mole of an ideal monoatomic gas is mixed with 3 mole of an ideal diatomic gas at room temperature then &gamma; for the mixture is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Total internal energy of gaseous mixture

U = U_{1} + U_{2} = (E_{t})_{1} + (E_{t} + E_{t})_{2}

= 2 (3 \times \frac{1}{2} RT) + 3 (3 \times \frac{1}{2} RT + 2 \times \frac{1}{2} RT) = \frac{21}{2} RT

N o w, C_{v} = \frac{1}{n} \cdot \frac{d U}{d T} = \frac{1}{5} \cdot \frac{d}{d T} (\frac{21}{2} RT) = \frac{21}{10} R

C_{P} = C_{V} + R = \frac{21}{10} R + R = \frac{31}{10} R

γ = \frac{C _{p}}{C _{v}} = \frac{31/10}{21/10} = 1.476

Q. 2 mole of an ideal monoatomic gas is mixed with 3 mole of an ideal diatomic gas at room temperature then γ for the mixture is____

Total internal energy of gaseous mixture U=U1​+U2​=(Et​)1​+(Et​+Et​)2​ =2(3×21​RT)+3(3×21​RT+2×21​RT)=221​RT Now,Cv​=n1​⋅dTdU​=51​⋅dTd​(221​RT)=1021​R CP​=CV​+R=1021​R+R=1031​R γ=Cv​Cp​​=21/1031/10​=1.476

Q. 2 mole of an ideal monoatomic gas is mixed with 3 mole of an ideal diatomic gas at room temperature then $γ$ for the mixture is____