Question

Calculate the work (kJ) done during a reaction in which the internal volume expands from 28 L to 51 L against an outside pressure of 4.9 atm.

Answer 1

Answer:

W= -11KJ

Explanation:

Given:

volume expands from 28 L to 51 L

pressure =4.9 atm.

We will need to Convert the pressure to Pascal SI

But 1 atm = 101,325 Pa.

Then,

Pressure= (4.9*101323)/1atm = 5*10^5 pa

Then we need to Convert the volumes to cubic meters

But we know that1 m³ = 1,000 L.

V1= 28L * 1m^3/1000L = 0.028m^3

V2=51L × 1m^3 /1000L =0.051m^3

The work done during the expansion of a gas can be calculated as

W= -P(V2-V1)

W= - 5*10^5(0.051m^3 - 0.028m^3)

W= -1.1× 10^4J

Then we can Convert the work to kiloJoule

But1 kJ = 1,000 J.

W= -1.1× 10^4J× 1kj/1000J

= -11KJ