Reaction on positive electrode (cathode):
PbSO₄₍s₎ + 2H₂O₍l₎ → 2e⁻ + PbO₂₍s₎ + 4H⁺₍aq₎ + SO₄²⁻₍aq₎.
s - solid.
l - liquid.
aq - dissolve in water.
PbSO4 is converted to Pb at one electrode (anode) and to PbO₂ at the other (cathode). Lead battery can be recharged, during recharging, an external source of energy is used.
I believe it’s D because they are the largest decimals
The final volume V₂=4.962 L
<h3>Further explanation</h3>
Given
T₁=20 + 273 = 293 K
P₁= 1 atm
V₁ = 4 L
T₂=100+273 = 373 K
P₂=780 torr=1,02632 atm
Required
The final volume
Solution
Combined gas law :
P₁V₁/T₁=P₂V₂/T₂
Input the value :
V₂=(P₁V₁T₂)/(P₂T₁)
V₂=(1 x 4 x 373)/(1.02632 x 293)
V₂=4.962 L