Question

11. A high altitude balloon contains 6.81 g of helium in 1.16 x 104 Lat –23°C. Assuming ideal gas behaviour, how many grams of helium would have to be added to increase the pressure to 4.0 x 10–3 atm?

Answer 1

here's your solution...

Answer 2

Answer:

Explanation:

we know that -

         

           $PV  =  nRT\\\\n=\dfrac{PV}{RT}$.......(i)

P $=4*10^{-3}$

V $=1.16*10^4$

R = 0.0821

T = 250 k

Put in eq. (i) ...

$n=\dfrac{4*10^{-3}*1.16*10^4}{0.0821*250}\\ \\n=\dfrac{46.4}{26.52}\\ \\n=2.26$

we also know that-

$n=\dfrac{w}{m}\\ \\2.26=\dfrac{w}{4}\\ \\w=2.26*4\\\\w=9.04$