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Question: Given, 1\(d{{m}^{3}}\) of an ideal gas at a pressure of 10 bar expands reversibly and isothermally t...

Given, 1dm3d{{m}^{3}} of an ideal gas at a pressure of 10 bar expands reversibly and isothermally to the final volume of 10 litre. What will be the heat absorbed in the process? (ln10=2.3)
(a) 1.15kJ
(b) 4.6kJ
(c) 2.3kJ
(d) 9.2kJ

Explanation

Solution

In this, first we have to find out the work done in the reversible isothermal expansion by using the formula as:
work done=−nRTln⁡V2V1  \begin{aligned} & work\text{ }done=-nRT\ln \dfrac{{{V}_{2}}}{{{V}_{1}}} \\\ & \\\ \end{aligned} and from this work done, we can easily calculate the heat released during the reversible isothermal expansion by applying the formula of first law of thermodynamics as: ΔU=q+w\Delta U=q+ w. Here, ΔT=0\Delta T=0 and so,ΔU=0\Delta U=0. Now solve it.

Complete answer:
By the reversible isothermal expansion, we mean the increase in the volume of the gas as it expands slowly at constant temperature and work done in reversible isothermal expansion can be calculated as;
work done=−nRTln⁡V2V1  \begin{aligned} & work\text{ }done=-nRT\ln \dfrac{{{V}_{2}}}{{{V}_{1}}} \\\ & \\\ \end{aligned} -----------------(1)
As we know that from the ideal gas equation that;
PV=nRTPV=nRT
Put the value of nRT in equation (1), it changes to as;
work done=−PVln⁡V2V1  \begin{aligned} & work\text{ }done=-PV\ln \dfrac{{{V}_{2}}}{{{V}_{1}}} \\\ & \\\ \end{aligned}------------------(2)
As;
The volume V1{{V}_{1}} = 1dm3d{{m}^{3}}=1L 9 (given)
The volume V2{{V}_{2}}=10L (given)
and pressure = 10 bar (given)
then, put all these values in equation (2), we get work done as;
work done=−1×10ln⁡101  = -10 ln10  = -10×2.3 (ln 10= 2.3)  = -23 L-bar \begin{aligned} & work\text{ }done=-1\times 10\ln \dfrac{10}{1} \\\ & \text{ = -10 ln10} \\\ & \text{ = -10}\times \text{2}\text{.3 (ln 10= 2}\text{.3)} \\\ & \text{ = -23 L-bar} \\\ \end{aligned}
Now, we will convert this work done in joule.
1 L-bar =100 J
And 1J= 1000 kJ
Then , the work done is;
work done=−23×1001000  = -2.3 kJ \begin{aligned} & work\text{ }done=-\dfrac{23\times 100}{1000} \\\ & \text{ = -2}\text{.3 kJ} \\\ \end{aligned}
So, from this, we can easily calculate the heat released by applying the first law of thermodynamics as;
ΔU=q+w\Delta U=q+ w
Since, the work is done isothermally, so temperature will be constant i.e. ΔT=0\Delta T=0 and so,ΔU=0\Delta U=0 .
Then, heat released is;
0=q+w q=2.3kJ \begin{aligned} & 0=q+ w \\\ & q=2.3kJ \\\ \end{aligned}
Hence, 1dm3d{{m}^{3}} of an ideal gas at a pressure of 10 bar when expands reversibly and isothermally to a final volume of 10 litre, the heat absorbed in the process is 2.3 kJ.

So, option (c) is correct.

Note:
The change in internal energy and temperature are interrelated to each other because as temperature increases internal energy of the system also increases and vice- versa and is constant when temperature becomes constant.