Calculate the second volumes. 955 L at 58 C and 108 KPa to 76 C and 123 KPa

The second volume :   V₂= 884.14 L

Given

955 L at 58 C and 108 KPa

76 C and 123 KPa

Required

The second volume

Solution

T₁ = 58 + 273 = 331 K

P₁ = 108 kPa

V₁ = 955 L

P₂ = 123 kPa

T₂ = 76 + 273 = 349

Use combine gas law :

P₁V₁/T₁ = P₂V₂/T₂

Input the value :

108 x 955/331 = 123 x V₂/349

V₂= 884.14 L

The question is incomplete, here is the complete question:

When a lead acid car battery is recharged by the alternator, it acts essentially as an electrolytic cell in which solid lead(II) sulfate PbSO₄ is reduced to lead at the cathode and oxidized to solid lead(II) oxide PbO at the anode.

Suppose a current of 96.0 A is fed into a car battery for 37.0 seconds. Calculate the mass of lead deposited on the cathode of the battery. Round your answer to 3 significant digits. Also, be sure your answer contains a unit symbol.

The mass of lead deposited on the cathode of the batter is 3.81 grams

Explanation:

To calculate the charge deposited, we use the equation:

We are given:

Current supplied = 96.0 A

Time = 37.0 seconds

Putting values in above equation, we get:

We know that:

96500 C of charge deposition is contained in 1 mole of electrons

So, 3552 C of charge deposition will be contained in = of electrons

The half reaction follows:

At cathode:  

Number of electrons exchanged are 2

So, moles of lead deposited =

To calculate the number of moles, we use the equation:

Molar mass of lead = 207.2 g/mol

Moles of lead = 0.0184 moles

Putting values in above equation, we get:

Hence, the mass of lead deposited on the cathode of the batter is 3.81 grams