$2$ moles of an ideal gas expands at a constant temperature of $27 C$ from $10$ litre to $100$ litre. The work done by the gas in this process will be nearly:

114800 J

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11486 J

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1148 J

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114.8 J

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Solution

The correct option is B $11486 J$
Given,

No. of moles, $n = 2$

Temperature, $T = 300 K$

$V_{i} = 10 l$

$V_{f} = 100 l$

For an isothermal process, work done is

$W = n R T ln (\frac{V_{f}}{V_{i}})$

$W = 2 \times 8.314 \times 300 l n (\frac{100}{10})$

$W = 11486.2 J$

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2 moles of an ideal gas expands at a constant temperature of 27C from 10 litre to 100 litre. The work done by the gas in this process will be nearly:

The correct option is B 11486JGiven, No. of moles, n=2 Temperature, T=300K Vi=10l Vf=100l For an isothermal process, work done is W=nRTln(VfVi) W=2×8.314×300ln(10010) W=11486.2J

$2$ moles of an ideal gas expands at a constant temperature of $27 C$ from $10$ litre to $100$ litre. The work done by the gas in this process will be nearly:

The correct option is B $11486 J$
Given,

No. of moles, $n = 2$

Temperature, $T = 300 K$

$V_{i} = 10 l$

$V_{f} = 100 l$

For an isothermal process, work done is

$W = n R T ln (\frac{V_{f}}{V_{i}})$

$W = 2 \times 8.314 \times 300 l n (\frac{100}{10})$

$W = 11486.2 J$