Question

One mole of an ideal monoatomic gas initially present in an insulated piston at 500K is expanded reversibly from 10 L to 40 L. Determine the workdone by the system.

Answer 1

Answer:

PLEASE MARK IT AS A BRAINLIEST ANSWER

Explanation:

the work done = -2.303nRTlog(V2/V1)

=-2.303×1×8.3×500×log(4)

=-5753.12JmolK

-5.753kJmolK

Answer 2

The workdone by system is -5.76 kJ.

Step-by-step explanation:

Given:

number of moles,n = 1

temperature = 500K

initial volume = 10 L

final volume = 40 L

To find: the work done by the system

Formula to be used:  $W=-2.303nRTlog\frac{V_{2}}{V_{1}}$

Step 1 of 1

The workdone for an isothermal reversible process is calculated by using the formula
$W=-2.303nRTlog\frac{V_{2}}{V_{1}}$

The value of gas constant, R that will be used here is 8.314 J/K mol.

Now substituting all the values given:

$W=-2.303*1*8.314*500log\frac{40}{10}\\\\W=-9,573.571*log4\\\\W=-9,573.571*0.602\\\\W=-5,763.864J\\\\W=-5.76kJ$

This is the work-done by the system.