Kinetic Theory

JEE Main level, single correct, medium

99 of 168

Question 99 of 168

Two moles of an ideal gas, with CPCV=53\dfrac{{C}_{P}}{{C}_{V}}=\dfrac{5}{3}, are mixed with three moles of another ideal gas CPCV=43\dfrac{{C}_{P}}{{C}_{V}}=\dfrac{4}{3}. The value of CPCV\dfrac{{C}_{P}}{{C}_{V}} for the mixture is

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Answer: (D) 1.421.42

Let n1,n2n_1,n_2 be the number of moles of gases 11 and 22.
We have Cp1Cv1=53\dfrac{C_{p_1}}{C_{v_1}}=\dfrac{5}{3} and We know Cp1=Cv1+RC_{p_1}=C_{v_1}+R
∴5Cv13=Cv1+R,2Cv13R1,Cv1=3R2\therefore \dfrac{5C_{v_1}}{3}=C_{v_1}+R,\dfrac{2C_{v_1}}{3}R_1,C_{v_1}=\dfrac{3R}{2}
AA and Cp1=Cv1+R=5R2C_{p_1}=C_{ v_ 1 }+R=\dfrac{5R}{2}
Also, Cp2Cv2=43\dfrac{C_{ p_ 2}}{C_{ v_ 2 }}=\dfrac{4}{3} and
Cp2=Cv2+RC_{p_2}=C_{v_2}+R
∴4Cv23=Cv2+R,Cv23=R,Cv2=3R\therefore \dfrac{4C_{ v_ 2 }}{3}=C_{ v_ 2 }+R, \dfrac{C_{ v_ 2 }}{3}=R,C_{ v_ 2 }=3R
Cp2=Cv2+R=4RC_{ p_ 2}=C_{ v_ 2 }+ R=4R
Now Cvmix=n1Cv1+n2Cv2n1+n2=2×3R2+3×3R2+3C_{ v_ {mix} }=\dfrac{n_1C_{ v_ 1 }+n_2C_{ v_ 2 }}{n_1+n_2}=\dfrac{2\times \dfrac{3R}{2}+3\times 3R}{2+3}
=3R+9R5=12R5=\dfrac{3R+9R}{5}=\dfrac{12R}{5}
Cpmix=n1Cp1+n2Cp2n1+n2=2×5R2+3×4R2+3C_{ p_ {mix} }=\dfrac{n_1C_{ p_ 1 }+n_2C_{ p_ 2 }}{n_1+n_2}=\dfrac{2\times \dfrac{5R}{2}+3\times 4R}{2+3}
=17R5=\dfrac{17R}{5}
lmix=CpmixCvmix=1712=1.4166=1.42l_{mix}=\dfrac{C_{ p_ {mix} }}{C_{ v_ {mix} }}=\dfrac{17}{12}=1.4166=1.42.

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