### Check Your Understanding

Line 1, $\text{\Delta}{E}_{\text{int}}=40\phantom{\rule{0.2em}{0ex}}\text{J;}$ line 2, $W=50\phantom{\rule{0.2em}{0ex}}\text{J}$ and $\text{\Delta}{E}_{\text{int}}=40\phantom{\rule{0.2em}{0ex}}\text{J;}$ line 3, $Q=80\phantom{\rule{0.2em}{0ex}}\text{J}$ and $\text{\Delta}{E}_{\text{int}}=40\phantom{\rule{0.2em}{0ex}}\text{J;}$ and line 4, $Q=0$ and $\text{\Delta}{E}_{\text{int}}=40\phantom{\rule{0.2em}{0ex}}\text{J}$

So that the process is represented by the curve $p=nRT\text{/}V$ on the *pV* plot for the evaluation of work.

$1.26\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{J}\text{.}$

### Conceptual Questions

Yes, as long as the work done equals the heat added there will be no change in internal energy and thereby no change in temperature. When water freezes or when ice melts while removing or adding heat, respectively, the temperature remains constant.

If more work is done on the system than heat added, the internal energy of the system will actually decrease.

Isothermal processes must be slow to make sure that as heat is transferred, the temperature does not change. Even for isobaric and isochoric processes, the system must be in thermal equilibrium with slow changes of thermodynamic variables.

Typically ${C}_{p}$ is greater than ${C}_{V}$ because when expansion occurs under constant pressure, it does work on the surroundings. Therefore, heat can go into internal energy and work. Under constant volume, all heat goes into internal energy. In this example, water contracts upon heating, so if we add heat at constant pressure, work is done on the water by surroundings and therefore, ${C}_{p}$ is less than ${C}_{V}$.

An adiabatic process has a change in temperature but no heat flow. The isothermal process has no change in temperature but has heat flow.

### Problems

$p\left(V-b\right)=\text{\u2212}{c}_{T}$ is the temperature scale desired and mirrors the ideal gas if under constant volume.

$3.53\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{4}\phantom{\rule{0.2em}{0ex}}\text{J}$

a. $({p}_{1}+3{V}_{1}^{2})({V}_{2}-{V}_{1})-3{V}_{1}({V}_{2}^{2}-{V}_{1}^{2})+({V}_{2}^{3}-{V}_{1}^{3})$; b. $\frac{3}{2}({p}_{2}{V}_{2}-{p}_{1}{V}_{1})$; c. the sum of parts (a) and (b); d. ${T}_{1}=\frac{{p}_{1}{V}_{1}}{nR}$ and ${T}_{2}=\frac{{p}_{2}{V}_{2}}{nR}$

a.

;

b. $W=4.39\phantom{\rule{0.2em}{0ex}}\text{kJ,}\phantom{\rule{0.2em}{0ex}}\text{\Delta}{E}_{\text{int}}=\mathrm{-4.39}\phantom{\rule{0.2em}{0ex}}\text{kJ}$

An adiabatic expansion has less work done and no heat flow, thereby a lower internal energy comparing to an isothermal expansion which has both heat flow and work done. Temperature decreases during adiabatic expansion.

### Additional Problems

a. ${W}_{AB}=0,{W}_{BC}=2026\phantom{\rule{0.2em}{0ex}}\text{J},{W}_{AD}=810.4\phantom{\rule{0.2em}{0ex}}\text{J},{W}_{DC}=0;$ b. $\text{\Delta}{E}_{AB}=3600\phantom{\rule{0.2em}{0ex}}\text{J},\text{\Delta}{E}_{BC}=374\phantom{\rule{0.2em}{0ex}}\text{J;}$ c. $\text{\Delta}{E}_{AC}=3974\phantom{\rule{0.2em}{0ex}}\text{J};$ d. ${Q}_{ADC}=4784\phantom{\rule{0.2em}{0ex}}\text{J;}$ e. No, because heat was added for both parts *AD* and *DC*. There is not enough information to figure out how much is from each segment of the path.

$2.4\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{J}$

A cylinder containing three moles of nitrogen gas is heated at a constant pressure of 2 atm. a. *−*1220 J; b. +1220 J

a. 7.6 L, 61.6 K; b. 81.3 K; c. $3.63\phantom{\rule{0.2em}{0ex}}\text{L}\xb7\text{atm}=367\phantom{\rule{0.2em}{0ex}}\text{J}$; d. −367 J

### Challenge Problems

a. 2.2 mol; b. ${V}_{A}=2.6\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-2}}{\text{m}}^{3}$, ${V}_{B}=7.4\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-2}}{\text{m}}^{3}$; c. ${T}_{A}=1220\phantom{\rule{0.2em}{0ex}}\text{K},{T}_{B}=430\phantom{\rule{0.2em}{0ex}}\text{K}$; d. 30,500 J