Question

26. a The work done in ergs for a reversible expansion of one mole of an ideal gas from a volume of 10 L to 20 L at 25°C is : (1)-2.303 X8.31 x 107 x 298 log2 (2) -2.303 0.0821 x 298 log2 (3) -2.303 0.0821 x 298 log 0.5 (4) -2.303 x2 x 298 log2​

Answer 1

Answer:

The work done during an isothermal reversible expansion of an ideal gas from one volume to another volume is option $(\ 1\ )$,

W   $=$  $-\ 2.303$  × $8.31$ × $10^{7}$ × $298$ × $log\ 2$

Explanation:

It is given that the process is a reversible expansion of one mole of an ideal gas· Here, the temperature is a constant one which is $27^{o}c$· So, the actual process is an isothermal reversible expansion of an ideal gas·

The equation for the work done during an isothermal reversible expansion of an ideal gas from one volume to another volume is given as,

    $W\ =\ -\ 2.303\ \ n\ R\ T\ \ log\ \frac{V_2}{V_1}$

where,

W  $=$  work done during the expansion

n   $=$  No of moles of the gas

R  $=$   Universal gas constant

T  $=$  Temperature

$V_1$  $=$  The initial volume of the gas

$V_2$ $=$  The final volume of the gas

Given that,

n  $=$  $1$ mol

T  $=$  $25^oc\ \ =\ 25\ +\ 273\ =\ 298\ K$

$V_1$ $=$  $10\ L$

$V_2$ $=$  $20\ L$

Since we need to find out the work done in ergs, the SI unit of the R-value becomes,

R   $=$  $8.314$ $J\ K^-^1\ mol^-^1$

where,

$1\ J\ =\ 10^7 ergs$

Therefore,

R   $=$  $8.314\ *\ 10^7$ $erg\ K^-^1\ mol^-^1$

On substituting these values we get,

W   $=$  $-\ 2.303$  × $1$ × $8.31$ × $10^{7}$ × $298$ × $log\ \frac{20}{10}$

W   $=$  $-\ 2.303$  × $8.31$ × $10^{7}$ × $298$ × $log\ 2$

Therefore, the work done during an isothermal reversible expansion of an ideal gas  from one volume to another volume is,

W   $=$  $-\ 2.303$  × $8.31$ × $10^{7}$ × $298$ × $log\ 2$