A gas is confined to a container with a massless piston at the top. A massless w

Question

A gas is confined to a container with a massless piston at the top. A massless wire is attached to the piston. When an external pressure of 2.00 is applied to the wire, the gas compresses from 5.90 to 2.95 . When the external pressure is increased to 2.50 , the gas further compresses from 2.95 to 2.36. In a separate experiment with the same initial conditions, a pressure of 2.50 was applied to the gas, decreasing its volume from 5.90 to 2.36 in one step. If the final temperature was the same for both processes, what is the difference between for the two-step process and for the one-step process in joules? Express your answer with the appropriate units.

Explanation / Answer

Remember that DELTA U = q + w Because in the case both parts DELTA U are the same thing we can state that: q1 + w1 = q2 + w2 with q1 and w1 belonging to the one step problem and q2 and w2 belonging to the second. This allows us to use simple algebra rules to state that: w1 - w2 = q2 - q1 We then use the formula: w = -P(DELTA V) on the single step answer (the second one): w1 = -P(DELTA V) : where P is the pressure and DELTA V is the difference in volumes. w1 = -2.50atm (2.16L - 5.40L) w1 = 8.10L atm To convert to Joules we use: 1 L atm = 101 Joules So 8.10L atm = 818.1 J Then repeat the process for the two step process w2 = Step 1 + Step 2 w2 = -P(DELTA V)1 + -P(DELTA V)2 w2 = -2.00atm (2.70L - 5.40L) + -2.50atm (2.16L - 2.70L) w2 = 5.4L atm + 1.35L atm w2 = 6.75L atm Convert to Joules 6.75L atm = 681.75 J Then go back to the forumula stated at the top w1 - w2 = q2 - q1 818.10 J - 681.75J = q2 - q1 136.35 J = q2 - q1

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