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88P

Expert-verifiedFound in: Page 582

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**An ideal gas initially at ${\mathbf{300}}{\text{\hspace{0.17em}}}{\mathbf{K}}{\mathbf{}}$****is compressed at a constant pressure of ${\mathbf{25}}{\mathbf{}}{\text{\hspace{0.17em}}}{\mathbf{N}}{\mathbf{/}}{{\mathbf{m}}}^{{2}}{\mathbf{}}$ ****from a volume of $\mathbf{3}\mathbf{.}\mathbf{0}\text{\hspace{0.17em}}{\mathbf{m}}^{3}$****to a volume of ${\mathbf{1}}{\mathbf{.}}{\mathbf{8}}{\mathbf{}}{{\mathbf{m}}}^{{3}}{\mathbf{}}$****. In the process, ${\mathbf{75}}{\text{\hspace{0.17em}}}{\mathbf{J}}{\mathbf{}}$****is lost by the gas as heat. What are **

**(a) the change in internal energy of the gas and **

**(b) the final temperature of the gas?**

- The internal energy change in the gas is$\Delta {E}_{\mathrm{int}}=-45\text{J}$ .
- The temperature of the gas changes to$180\text{\hspace{0.17em}K}$ .

**The process that takes place at constant pressure is called an isobaric process. Using the formulas for work done W in an isobaric process, and the first law of thermodynamics, we can find the change in the internal energy ${{\mathbf{\Delta E}}}_{{\mathrm{int}}}$**** of the gas. Also, by using the ideal gas law in the ratio form, we can find the final temperature ${{\mathbf{T}}}_{{2}}$of the gas.**

**1.The work done W in anisobaric process is**

**${\mathbf{W}}{\mathbf{=}}{\mathbf{p\Delta V}}$**

**2.The ideal gas law in ratio form is**

**${{\mathbf{T}}}_{{2}}{\mathbf{=}}{{\mathbf{T}}}_{{1}}\left(\frac{{\mathbf{V}}_{2}}{{\mathbf{V}}_{1}}\right)$**

**3.The ****change in the internal energy is**

${{\mathbf{\Delta E}}}_{{\mathrm{int}}}{\mathbf{=}}{\mathbf{Q}}{\mathbf{-}}{\mathbf{W}}$

- The pressure .$p=25{\text{\hspace{0.17em}N/m}}^{\text{2}}$
- The final volume ${V}_{f}=1.8{\text{m}}^{3}$
- The initial volume .${V}_{i}=3.0{\text{m}}^{3}$
- The initial temperature ${T}_{1}=300\text{K}$
- In the process, the energy lost by the gas as heat is$Q=-75\text{J}$ .

a.The work done in anisobaricprocess is expressed as-

$W=p\Delta V$

Where p is the pressure and $\Delta V=({V}_{f}-{V}_{i})$is change in the volume

Therefore,

$\begin{array}{c}W=p({V}_{f}-{V}_{i})\\ =25\text{\hspace{0.17em}N}/{\text{m}}^{2}\times (1.8{\text{\hspace{0.17em}m}}^{3}-3.0{\text{\hspace{0.17em}m}}^{3})\\ =-30\text{\hspace{0.17em}J}\end{array}$

From the first law of thermodynamics, we have-

$\Delta {E}_{\mathrm{int}}=Q-W$

Where Q is the heat transferred and $\Delta {\text{E}}_{int}$ is the change in the internal energy o the gas. It is given that Q amount of energy leaves the system; therefore, we can consider Q as negative. $Q=-75\text{\hspace{0.17em}J}$

$\begin{array}{c}\Delta {E}_{\mathrm{int}}=-75\text{\hspace{0.17em}J}-(-30\text{\hspace{0.17em}J})\\ =-75\text{\hspace{0.17em}J}+\left(30\text{\hspace{0.17em}J}\right)\\ =-45\text{\hspace{0.17em}J}\end{array}$

Internal energy is decreased by $-45\text{\hspace{0.17em}J}$

b.Since the pressure is constant and the number of moles presumed are constant, the ideal gas law can be written as-

$\begin{array}{c}{T}_{2}={T}_{1}\left(\frac{{V}_{2}}{{V}_{1}}\right)\\ =300\text{K}\times \left(\frac{1.8\text{\hspace{0.17em}}{\text{m}}^{3}}{3.0\text{\hspace{0.17em}}{\text{m}}^{3}}\right)\\ =180\text{\hspace{0.17em}K}\end{array}$

The change in internal energy of the gas is $-45J$and the final temperature of the gas is $180\text{\hspace{0.17em}K}$.

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