Question

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

Answer 1

Answer:

1. -1,42x10³ J
2. -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!

Answer 2

Work = -Pressure external*Volume change 

-1 atm * (20 - 1) = -19 atm-L. 

1 atm-L = 101.324999971 J 

So -19 * 101.32 = -1925 J 

B. 

Calculate each step individually then sum the answers. 

For step 1) 
-5 * (4-1) = -15 atm-L 

step 2) 
-1 * (20-4) = -16 atm-L 

Thus total work = -31 atm-L 

convert to joules, 

-3141 J