Heat and Thermodynamics

Given:
Initial work done by the gas, W1=W
Final volume for the first process, Vf1=2V
Initial volume for the first process, Vi1=V
Temperature for the first process, T1=T
Number of moles for the first process, n=2
The work done by an ideal gas during an isothermal expansion is given by the equation:
W=n⋅R⋅T‌ln(‌

Vf

Vi

)
Substituting the given values for the first process:
W1=2⋅R⋅T‌ln(‌

2V

V

)=2⋅R⋅T‌ln‌2
For the second process:
Final volume, Vf2=8V
Initial volume, Vi2=V
Temperature, T2=‌

T

2

Number of moles, n2=4
The work done for the second process is:
W2=n2⋅R⋅T2‌ln(‌

Vf2

Vi2

)
Substituting the values:
‌W2=4⋅R⋅‌

T

2

‌ln(‌

8V

V

)
‌W2=2⋅R⋅T‌ln(23)
‌W2=3⋅2⋅R⋅T‌ln‌2
This simplifies to:
W2=3⋅W
Hence, the work done by the ideal gas in the second process is 3W.