TS EAMCET 11-Sep-2020 Shift 2 Solved Paper

Given, pressure of mixture (p0)=1.1atm
Molecular weight of two gases =4 and 40 .
∴ Number of moles of lighter gas =m/4 where, m= mass of gases
and, number of moles of heavier gas =‌

m

40

Thus, total number of moles =‌

m

4

+‌

m

40

=‌

11m

40

Hence, mole fraction of lighter gas
‌

m/4

11m/40

=‌

10

11

∴ Partial pressure of lighter gas
=p0×‌

10

11

=1.1×‌

10

11

=1atm