Question

A sample of air occupies 4.40 L when the pressure is 2.60 atm.

Answer 1

Explanation:

Given that,

Initial volume, V₁ = 4.40 L

Initial pressure, P₁ = 2.6 atm

(a) Final pressure, P₂ = 6.2 atm

As per the relation,

$P_1V_1=P_2V_2\\\\V_2=\dfrac{P_1V_1}{P_2}\\\\V_2=\dfrac{2.6\times 4.4}{6.2}\\\\V_2=1.84\ L$

(b) Again using the above relation,

$P_2=\dfrac{P_1V_1}{V_2}\\\\P_2=\dfrac{2.6\times 4.4}{0.0290 }\\\\P_2=394.48\ atm$

Hence, this is the required ssolution.