Suggested languages for you:

Americas

Europe

Q13P

Expert-verifiedFound in: Page 578

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**A sample of an ideal gas is taken through the cyclic process abca as shown in figure. The scale of the vertical axis is set by ${\text{p}}_{\text{b}}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{5}\text{\hspace{0.17em}}\mathbf{kPa}$ ****and ${{\text{p}}}_{{\text{ac}}}{\mathbf{=}}{\mathbf{2}}{\mathbf{.}}{\mathbf{5}}{\text{\hspace{0.17em}}}{\mathbf{kPa}}$****. At point a,${\text{T}}{\mathbf{=}}{\mathbf{200}}{\text{\hspace{0.17em}}}{\mathbf{K}}$ ****.**

**aHow many moles of gas are in the sample?**

**What are:**

**b.The temperature of gas at point b**

**c.Temperature of gas at point c**

**d.The net energy added to the gas as heat during the cycle?**

- The number of moles of gas present in the sample are $\text{1.5 mol}$.
- The temperature of the gas at point b is .$1.8\times {10}^{3}\text{K}$
- The temperature of the gas at point c is $6.0\times {10}^{2}\text{K}$.
- The net energy added to the gas as heat during the cycle is $5.0\times {10}^{3}\text{J}.$

${P}_{b}=7.5\text{kPa}=7500\text{Pa}\phantom{\rule{0ex}{0ex}}{P}_{ac}=2.5\text{\hspace{0.17em}kPa}=2500\text{\hspace{0.17em}Pa}$

- At point a, $T=200\text{\hspace{0.17em}K}$
- At point a, $V=1{\text{\hspace{0.17em}m}}^{3}$
- At point b, $V=3{\text{\hspace{0.17em}m}}^{3}$

**Find the**** number of moles and the temperature of the gas at points b and c present in the sample using the gas law. Net energy added can be found from the net work done by the gas using the given graph.**

Formula is as follow:

${p}_{i}{v}_{i}=nR{T}_{i}$

Here, *p* is pressure, *v* is volume, *T* is temperature, *R* is universal gas constant and *n* is number of moles.

According to the gas law,

$PV=nRT\phantom{\rule{0ex}{0ex}}n=\frac{PV}{RT}$

From the given graph, at point a,

$n=\frac{2500\left(1\right)}{8.314\left(200\right)}\phantom{\rule{0ex}{0ex}}n=1.5\text{\hspace{0.17em}mol}\phantom{\rule{0ex}{0ex}}$

Hence, the number of moles of gas present in the sample are .$\text{1.5 mol}$

By gas law,

$\frac{{P}_{b}{V}_{b}}{{P}_{a}{V}_{a}}=\frac{{T}_{b}}{{T}_{a}}$

${T}_{b}=\frac{{P}_{b}{V}_{b}}{{P}_{a}{V}_{a}}{T}_{a}\phantom{\rule{0ex}{0ex}}{T}_{b}=\frac{7500\left(3\right)}{2500\left(1\right)}\left(200\right)$

${T}_{b}=1.8\times {10}^{3}\text{K}$

Hence, the temperature of the gas at point b is $1.8\times {10}^{3}\text{K}$.

** **

Similarly,

${T}_{c}=\frac{{P}_{c}{V}_{c}}{{P}_{a}{V}_{a}}{T}_{a}\phantom{\rule{0ex}{0ex}}{T}_{c}=\frac{2500\left(3\right)}{2500\left(1\right)}\left(200\right)\phantom{\rule{0ex}{0ex}}$

${T}_{c}=600=6.0\times {10}^{2}\text{K}$

Hence, the temperature of the gas at point c is $6.0\times {10}^{2}\text{K}$.

** **

First law of thermodynamics gives,

$\u25b3E=Q-W$

For isothermal process,

$Q=W\phantom{\rule{0ex}{0ex}}Q=\text{Area under the curve}\phantom{\rule{0ex}{0ex}}Q=\frac{1}{2}\left(\text{ac}\right)\left(\text{bc}\right)$

Substitute the values as:

$Q=\frac{1}{2}\left(2\right)(7500-2500)\phantom{\rule{0ex}{0ex}}$

$Q=5000=5.0\times {10}^{3}\text{J}$

Hence, the net energy added to the gas as heat during the cycle is $5.0\times {10}^{3}\text{J}.$

Therefore, the number of moles of gas present in the sample,temperature at a point, and heat added can be foundusing gas law.

94% of StudySmarter users get better grades.

Sign up for free