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,
[tex]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[/tex]
(b) Again using the above relation,
[tex]P_2=\dfrac{P_1V_1}{V_2}\\\\P_2=\dfrac{2.6\times 4.4}{0.0290 }\\\\P_2=394.48\ atm[/tex]
Hence, this is the required ssolution.
