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Chapter 6: Problem 38

Calculate the internal energy change for each of the following. a. One hundred \((100 .)\) joules of work is required to compress a gas. At the same time, the gas releases 23 \(\mathrm{J}\) of heat. b. A piston is compressed from a volume of 8.30 \(\mathrm{L}\) to 2.80 \(\mathrm{L}\) against a constant pressure of 1.90 \(\mathrm{atm}\) . In the process, there is a heat gain by the system of 350. J. c. A piston expands against 1.00 atm of pressure from 11.2 \(\mathrm{L}\) to 29.1 \(\mathrm{L}\) . In the process, 1037 \(\mathrm{J}\) of heat is absorbed.

Short Answer

Expert verified
The internal energy changes for each process are: a. \( \Delta U = 77 \, J \) b. \( \Delta U = -678.085 \, J \) c. \( \Delta U = 2851.37 \, J \)

Step by step solution

01

a. Identify work and heat

In this case, work required to compress the gas is given as 100 J. This means that work is done on the gas, so w = -100 J. The gas releases 23 J of heat, so q = -23 J.
02

a. Calculate internal energy change

Using the first law of thermodynamics, ∆U = q - w = -23 - (-100) = 77 J.
03

b. Identify work and heat

Here, we need to calculate the work done by the system when the piston is compressed. We're given the initial volume V1 = 8.30 L, the final volume V2 = 2.80 L, and the constant pressure P = 1.90 atm. The heat gained by the system is 350 J.
04

b. Calculate work done by the system

The work done by the system is w = -P∆V. We need to first convert the pressure to J/L (1 atm = 101.3 J/L) and calculate the change in volume: ∆V = V2 - V1 = 2.80 - 8.30 = -5.50 L. Then, w = -1.90 * 101.3 * (-5.50) = 1028.085 J.
05

b. Calculate internal energy change

Now, we can use the first law of thermodynamics to find the change in internal energy: ∆U = q - w = 350 - 1028.085 = -678.085 J.
06

c. Identify work and heat

In this case, we have an expansion of the gas in the piston with an initial volume V1 = 11.2 L, a final volume V2 = 29.1 L, and an external pressure of 1.00 atm. The system absorbs 1037 J of heat.
07

c. Calculate work done by the system

We'll calculate the work: w = -P∆V. We already have the pressure in J/L and need to calculate the change in volume: ∆V = V2 - V1 = 29.1 - 11.2 = 17.9 L. Then, w = -1.00 * 101.3 * 17.9 = -1814.37 J.
08

c. Calculate internal energy change

Finally, we use the first law of thermodynamics to find the change in internal energy: ∆U = q - w = 1037 - (-1814.37) = 2851.37 J.

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Most popular questions from this chapter

Consider the dissolution of \(\mathrm{CaCl}_{2} :\) $$ \mathrm{CaCl}_{2}(s) \longrightarrow \mathrm{Ca}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q) \quad \Delta H=-81.5 \mathrm{kJ} $$ An 11.0 -g sample of \(\mathrm{CaCl}_{2}\) is dissolved in 125 g water, with both substances at \(25.0^{\circ} \mathrm{C}\) . Calculate the final temperature of the solution assuming no heat loss to the surroundings and assuming the solution has a specific heat capacity of 4.18 $\mathrm{J} /^{\prime} \mathrm{C} \cdot \mathrm{g} .$

For the following reactions at constant pressure, predict if $\Delta H>\Delta E, \Delta H<\Delta E,\( or \)\Delta H=\Delta E .$ a. \(2 \mathrm{HF}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{F}_{2}(g)\) b. $\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)$ c. $4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$

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