AP EAPCET 07-Jul-2022 Shift 1 Paper

For a gas mixture,
rmixture=

Cp(mix)

CV(mix)

=

( n1Cp1+n2Cp2 n1+n2 )

( n1CV1+n2CV2 n1+n2 )

⇒rmix=

n1Cp1+n2Cp2

n1CV1+n2CV2

...(i)
Now, in given mixture we have 4g of oxygen (O2) and 4g of helium (He)
Now, number of moles of oxygen,
n1=

m

M

=

Sample‌mass

Molar‌mass

⇒n1=

4

32

=

1

8

Similarly number of moles of helium, n2=

4

4

=1
Also oxygen is a diatomic gas so,
Cp1=

7

2

R and CV1=

5

2

R
Helium is monoatomic, hence
Cp2=

5

2

R,CV2=

3

2

R

Substituting these values in eq. (i) we get,
rmix=

1 8 × 7 2 R+1× 5 2 R

1 8 × 5 2 R+1× 3 2 R

=

7 16 + 5 2

5 16 + 3 2

or
rmix=

7+40

5+24

=

47

29