Consider an ideal gas enclosed in a 1.00 L container at an internal pressure of 18.0 atm. If the gas expands against a constant external pressure of 1.00 atm to a final volume of 10.0 L, what is the work done? Now calculate the work done if this process is carried out in two steps.
1. First, let the gas expand against a constant external pressure of 5.00 atm to a volume of 2.00 L.
2. From there, let the gas expand to 10.0 L against a constant external pressure of 1.00 atm.

A=-0.9117kJ.

B= -0.5065kJ

C=-0.8104KJ

Explanation: A s gas expands, it produces works . Therefore

Work done by gas expansion,

W  = P Δ V  

where P =external  pressure,

ΔV = change in volume (final - Initial)

W = -1 atm (10 - 1 ) L

W =  -9L. atm  x  101.3J/  1L. atm

W = 911.7Joules ≈ -0.9117kJ.

1.The work done in the first step is:

W= - 5.00 atm × (2.00 L - 1.00 L) = -5 atm·L ×(101.3J / 1 atm·L)

=506.5J = -0.5065kJ

2.The work done in the second step is:

w = - 1.00 atm × (10.0 L - 2.00 L) = -8atm·L ×(101.3J / 1 atm·L

=810.4J

w=-0.8104KJ

The work done in the whole  process is:

w =-0.5065kJ --0.8104KJ  = -1.3169KJ

answer : one can design an experiment using a liquid and an additive that can be added to the liquid and observe the chemical changes in the reaction species.

one would need a beaker which is filled with a known amount of the liquid in which thermometer is placed to measure the temperature changes; after addition of the additive to the liquid solution one can note down the temperature change and find whether the reaction is endothermic or exothermic in nature. by subtracting the initial temperature value with the final temperature one can find the temperature changes.

various combinations of liquid and additive can be used. liquid in this case can be a known quantity of water and additives can be epsom salt or without epsom salt the experiment can be conducted to notice the changes. water and ice, hydrogen peroxide and dry yeast, vinegar and baking soda,etc.

i think its 100

i hope this !