# This was a question from Pradeep book for Class 11 Physics. Is the question incomplete and requires more information? If no, please give me a solution. Thanks!

Then teach the underlying concepts
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Jane Share
Mar 9, 2018

Work done $W$ in case of isobaric process is $n R \delta T$

Given , $n = 1$ and $\delta T = {1}^{\circ} C = 1 K$ (as ${1}^{\circ}$ Change in Celcius scale is equals to $1 K$ change in Kelvin scale.)

And, $R = 8.31 J {K}^{-} 1 m o {l}^{-} 1$

So, $W = 1 \cdot 8.31 \cdot 1 = 8.31 J = \frac{8.31}{4.2} = 2.08$ Calories

Remember the following formulae for thermodynamics for quick answering,where $\partial Q$ stands for heat energy required, $\partial U$ stands for change in internal energy and $\partial W$ stands for work done.

And in the given expressions, ${V}_{2}$ is the final volume and ${V}_{1}$ is the initial volume, ${P}_{1}$ is initial pressure and ${P}_{2}$ is final pressure,and $\delta T$ is change in temperature.

$\gamma$ stands for ${C}_{p} / {C}_{v}$

For Isothermal process

$\partial Q = n R T \ln \left(\frac{{V}_{2}}{{V}_{1}}\right)$

$\partial U = 0$

$\partial W = \partial Q = n R T \ln \left(\frac{{V}_{2}}{{V}_{1}}\right) = n R T \ln \left(\frac{{P}_{2}}{{P}_{1}}\right)$ (as $P V = c o n s \tan t$)

$\partial Q = 0$

$\partial U = n {C}_{v} \delta T = \frac{{P}_{2} {V}_{2} - {P}_{1} {V}_{1}}{\gamma - 1}$

$\partial W = - \partial U = - n {C}_{v} \delta T = \frac{{P}_{1} {V}_{1} - {P}_{2} {V}_{2}}{\gamma - 1}$

For isobaric process

$\partial Q = n {C}_{p} \delta T$

$\partial U = n {C}_{v} \delta T$

$\partial W = n R \delta T$

For isometric process

$\partial Q = n {C}_{v} \delta T$

$\partial U = \partial Q = n {C}_{v} \delta T$

$\partial W = 0$

Remember $T$ is in Kelvin scale and don't forget to apply sign conventions

Also remember, ${C}_{p} = {C}_{v} + R$ and $\gamma = 1 + \frac{2}{f}$ where $f$ is the degrees of freedom

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