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Chapter 8: Problem 1146

Work done per mole in an isothermal ? change is (A) RT log \(_{e}\left(\mathrm{v}_{2} / \mathrm{v}_{1}\right)\) (B) \(\mathrm{RT} \log _{10}\left(\mathrm{v}_{2} / \mathrm{v}_{1}\right)\) (C) RT \(\log _{10}\left(\mathrm{v}_{1} / \mathrm{v}_{2}\right)\) (D) RT \(\log _{\mathrm{e}}\left(\mathrm{v}_{1} / \mathrm{v}_{2}\right)\)

Short Answer

Expert verified
The short answer is: \[W = RT \log _{e}\left(\frac{V_2}{V_1}\right)\]

Step by step solution

01

Recall the ideal gas law

The ideal gas law is given by: \[PV = nRT\] Where P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant, and T is the temperature. Since we're given an isothermal process, the temperature (T) won't change.
02

Express work done in an isothermal process

The work done (W) in an isothermal process can be expressed as: \[W = -nRT \log_e \left(\frac{V_2}{V_1}\right)\]
03

Comparing the given options with the work formula

Comparing the formula with the given options, we can see that the option (A) has the same expression as the work formula.
04

Conclusion

Based on the comparison, we can conclude that the work done per mole in an isothermal process is: (A) RT log\(_e\left(\mathrm{v}_{2} / \mathrm{v}_{1}\right)\).

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Most popular questions from this chapter

Carnot engine working between \(300 \mathrm{~K}\) and \(600 \mathrm{~K}\) has work output of \(800 \mathrm{~J}\) per cycle. What is amount of heat energy supplied to the engine from source per cycle (A) \(1600\\{\mathrm{~J} /\) (cycle)\\} (B) \(2000\\{\mathrm{~J} /(\mathrm{cycle})\\}\) (C) \(1000\\{\mathrm{~J} /(\) cycle \()\\}\) (D) \(1800\\{\mathrm{~J} /(\) cycle \()\\}\)

An ideal gas at \(27 \mathrm{C}\) is Compressed adiabatically, to \(\\{8 / 27\\}\) of its original Volume. If \(\mathrm{v}=(5 / 3)\), then the rise in temperature is (A) \(225 \mathrm{k}\) (B) \(450 \mathrm{~K}\) (C) \(375 \mathrm{~K}\) (D) \(405 \mathrm{~K}\)

For an adiabatic expansion of a perfect gas, the value of $\\{(\Delta \mathrm{P}) / \mathrm{P}\\}$ is equal to (A) \(-\sqrt{\gamma}\\{(\Delta \mathrm{r}) / \mathrm{v}\\}\) (B) \(-\\{(\Delta \mathrm{v}) / \mathrm{v}\\}\) (C) \(-\gamma^{2}\\{(\Delta \mathrm{v}) / \mathrm{v}\\}\) (D) \(-\gamma\\{(\Delta \mathrm{v}) / \mathrm{v}\\}\)

Water of volume 2 liter in a container is heated with a coil of $1 \mathrm{kw}\( at \)27^{\circ} \mathrm{C}$. The lid of the container is open and energy dissipates at the rate of \(160(\mathrm{~J} / \mathrm{S}) .\) In how much time temperature will rise from \(27^{\circ} \mathrm{C}\) to $77^{\circ} \mathrm{C}\(. Specific heat of water is \)4.2\\{(\mathrm{KJ}) /(\mathrm{Kg})\\}$ (A) \(7 \mathrm{~min}\) (B) \(6 \min 2 \mathrm{~s}\) (C) \(14 \mathrm{~min}\) (D) \(8 \min 20 \mathrm{~S}\)

A gas expands \(0.25 \mathrm{~m}^{3}\) at Constant Pressure \(10^{3}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\) the work done is (A) \(250 \mathrm{~J}\) (B) \(2.5\) erg (C) \(250 \mathrm{~W}\) (D) \(250 \mathrm{~N}\)
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