Ncert Exercises & Solutions

For an ideal gas, the work of reversible expansion under isothermal condition can be calculated by using the expression $W = - nRT \ln \frac{V_{f}}{V_{i}}$ . A sample containing 1.0 mol of an ideal gas is expanded isothermally and reversible to ten times of its original volume, in to separate experiments. The expansion is carried out at 300 K and at 600 K respectively. Choose the correct option.

1. Work done at 600 K is 20 times the work done at 300 K

2. Work done at 300 K is twice the work done at 600 K

3. Work done at 600 K is twice the work done at 300 K

4. $∆ U = 0$ in both cases

(c, d)

Given that, the work of reversible expansion under isothermal condition can be calculated by using the expression

W = - nRT \ln \frac{V_{f}}{V_{i}}

V_{f} = 10 V_{i}

T_{2} = 600 K

T_{1} = 300 K

Putting these values in above expression

W_{600 K} = 1 \times R \times 600 K \ln \frac{10}{1}

W_{300 K} = 1 \times R \times 300 K \ln \frac{10}{1}

Ration = \frac{W_{600 K}}{W_{300 K}} = \frac{1 \times R \times 600 K \ln \frac{10}{1}}{1 \times R \times 300 K \ln \frac{10}{1}} = \frac{600}{300} = 2

For isothermal expansion of ideal gases,

∆ U = 0

. Since, temperature is constant this means there is no change is internal energy.

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