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

8 m/s because for him to get the same amount of energy as ray is putting out, bobby, who weighs half as much as ray, needs to run twice as fast.

the answer is na (sodium)