Consider an ideal gas enclosed in a 1.00 l container at an internal pressure of 10.0 atm.?
consider an ideal gas enclosed in a 1.00 l container at an internal pressure of 10.0 atm.

calculate the work, w, if the gas expands against a constant external pressure of 1.00 atm exerted by the surroundings to a final volume of 15.0 l.

now calculate the work done if this process is carried out in two steps: step 1: first let the gas expand against a constant external pressure of 5.00 atm to a volume of 3.00 l. step 2: now let the gas expand to 15.0 l against a constant external pressure of 1.00

-1,42x10³ J-2,23x10³ J

Explanation:

When a gas expands it produce energy as work. The equation that explain this phenomenon is:

Work (w) = - External pressure × ΔV

In the first question. We have an external pressure of 1,00 atm and the ΔV is:

1,00 L - 15,0 L = 14 L

Thus, in this case work is:

w = - 1,00 atm × 14,0 L = -14,0 atm·L

The problem does not especify the units to solve this problem, for international system of units (SI) Joules is the correct unit for energy, so:

-14,0 atm·L ×(101,325 J / 1 atm·L) = -1,42x10³ J

In the second problem occurs the same principle but occurs in two steps. The work done in the first step is:

w = - 5,00 atm × (3,00 L - 1,00 L) = -10 atm·L ×(101,325 J / 1 atm·L) =

-1,01x10³ J

In second step:, the work is:

w = - 1,00 atm × (15,0 L - 3,00 L) = -12 atm·L ×(101,325 J / 1 atm·L) =

-1,22x10³ J

Thus, the work done in the process is:

w = -1,01x10³ J - 1,22x10³ J = -2,23 x10³ J

I hope it helps!

Ibelieve the answer is d

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