Question

One mole of an ideal gas expands isothermally to double its volume at 27°C. Then the work done by
the gas is nearly​

Answer 1

Answer:

$\boxed{\mathfrak{Work \ done \ by \ the \ gas = 300R \: log_{e}2}}$

Explanation:

Let initial volume ($\rm V_1$) be 'V'

Final volume ($\rm V_2$) will be = 2V

Temperature (T) = 27°C = 300 K

Number of moles of gas (n) = 1

Work done (W) in isothermal proces is given as:

$\boxed{ \bold{W = nRT \: log_{e}(\dfrac{V_2}{V_1})}}$

R → Gas Constant

By substituting values in the equation we get:

$\rm \implies W = nRT \: log_{e}(\dfrac{V_2}{V_1}) \\ \\ \rm \implies W = 1 \times 300R \: log_{e}(\dfrac{2V}{V}) \\ \\ \rm \implies W = 300R \: log_{e}2$

By substituting different values of gas constant you can get different values of work done in different units.

Like,

R → 2 cal/mol-K

$\rm \implies W = 300 \times 2 \times 0.693 \\ \\ \rm \implies W = 415.8 \ cal$

R → 8.314 J/mol-K

$\rm \implies W = 300 \times 8.314 \times 0.693 \\ \\ \rm \implies W = 1728.48 \ J$

Answer 2

Answer:

Work done is $300Rlog_{e} 2$.

Explanation:

As per the information given in the question,

One mole of an ideal gas expands isothermally to double its volume.

This task is done at $27^{0}C$

Assuming initial volume be $V$

Assuming initial volume be $2V$

Temperature is given as $T=27^{0} C$

Work done in isothermal process is given by:

$W=nRTlog_{e} \frac{V_{2} }{V_{1} }$

$R$ is gas constant

Using given vales, we get

$W=nRTlog_{e} \frac{V_{1} }{V_{2} }$

$W=nRTlog_{e} \frac{2V }{V }$

$W=nRTlog_{e} 2$

Temperature in Kelvin is $273+27=300K$

$n=1$

Hence, work is $300Rlog_{e} 2$

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