Question

2 mole of a ideal gas at 2 bar and 27 deg celcius expands isothermally against a constant pressure of 1 bar. the work done by the gas is equal to

Answer 1

WB=Integral(p dv)
WB=Integral(nRT/V)(dv)
WB=(n)(R)(T)(In (v2/v1))
(p2)(v2)=(p1)(v1) since ideal gas and isothermal
v2/v1=p1/p2
WB=(n)(R)(T)(In(p1/p2))
WB=(2)(8.314J/gmol^-k)(300.2)(In(2/1)
WB=3460N-m=3460J

Answer 2

Answer : The maximum work done is, -11.52 J

Solution :

This is the case of isothermal reversible expansion of gas.

Formula used :

$w=-2.303nRT\log (\frac{P_1}{P_2})$

where,

w = work done by the gas  = ?

n = number of moles of gas  = 2 mole

R = gas constant = 8.314 J/mole K

T = temperature of gas = $27^oC=273+27=300K$

$P_1$ = initial pressure of gas = 2 bar

$P_2$ = final pressure of gas = 1 bar

Now put all the given values in the above formula, we get the work done.

$w=-2.303\times 2mole\times 8.314J/moleK\times 300K\times \log (\frac{2bar}{1bar})$

$w=-11.52J$

Therefore, the maximum work done is, -11.52 J