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Standard XIII

Chemistry

Work Done in Isothermal Reversible Process

How much work...

Question

How much work in $k J m o l^{- 1}$ unit is done by reversible and isothermal expansion of $1.2$ mole of an ideal gas to $10$ times of its original volume at $27^{o} C$ ?

4.191

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6.892

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- 6.892

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- 4.191

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Solution

The correct option is C $- 6.892$
$W = - 2.303 n R T l o g \frac{V_{2}}{V_{1}} \Rightarrow - 2.303 \times 1.2 \times 8.314 \times 10^{- 3} \times 300 l o g \frac{10 V_{1}}{V_{1}} \Rightarrow - 6.892 k J m o l^{- 1}$

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Q. How much work in

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Q. How much work in

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For an ideal gas, the work of reversible expansion under isothermal condition can be calculated by using the expression $W = - n R T l n \frac{V_{f}}{V_{i}}$ . A sample containing 1.0 mol of an ideal gas is expanded isothermally and reversible to ten times of its original volume, in two separate experiments. The expansion is carried out at 300 K and at 600 K respectively. Choose the correct option.
(a) Work done at 600 K is 20 times the work done at 300 K
(b) Work done at 300 K is twice the work done at 600 K
(c) Work done at 600 K is twice the work done at 300 K
(d) $Δ$ U = 0 in both cases

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How much work in kJ mol−1 unit is done by reversible and isothermal expansion of 1.2 mole of an ideal gas to 10 times of its original volume at 27oC ?

The correct option is C −6.892W=−2.303n RT logV2V1⇒−2.303×1.2×8.314×10−3×300 log10V1V1⇒−6.892 kJ mol−1

How much work in $k J m o l^{- 1}$ unit is done by reversible and isothermal expansion of $1.2$ mole of an ideal gas to $10$ times of its original volume at $27^{o} C$ ?

The correct option is C $- 6.892$
$W = - 2.303 n R T l o g \frac{V_{2}}{V_{1}} \Rightarrow - 2.303 \times 1.2 \times 8.314 \times 10^{- 3} \times 300 l o g \frac{10 V_{1}}{V_{1}} \Rightarrow - 6.892 k J m o l^{- 1}$