JEE MAIN - Physics (2022 - 27th June Morning Shift - No. 19)
A mixture of hydrogen and oxygen has volume 2000 cm3, temperature 300 K, pressure 100 kPa and mass 0.76 g. The ratio of number of moles of hydrogen to number of moles of oxygen in the mixture will be:
[Take gas constant R = 8.3 JK$$-$$1mol$$-$$1]
$${1 \over 3}$$
$${3 \over 1}$$
$${1 \over 16}$$
$${16 \over 1}$$
Објашњење
$${P_1}V = {n_1}RT$$
$${P_2}V = {n_2}RT$$
$$\Rightarrow$$ (100 kPa) V = (n1 + n2)RT
$$ \Rightarrow {n_1} + {n_2} = {{(100\,kPa)(2000\,c{m^3})} \over {8.3 \times 300}}$$ ..... (1)
Also, n1 $$\times$$ 2 + n2 $$\times$$ 32 = 0.76 ...... (2)
Solving (1) and (2),
n1 = 0.06
n2 = 0.02
$$ \Rightarrow {{{n_1}} \over {{n_2}}} = 3$$