the volume of a monoatomic ideal gas doubles in an adiabatic expansion. by considering 135 moles of gas with an initial pressure of 330kPa and an initial volume of 1.2m^3, the gas is then expands adiabatically to a volume of 2.4m^3.

FIND THE WORK DONE BY THE GAS.

Thank you.

γ=(i+2)/i=5/3=1.67

W=νRT₁ /(γ-1)}•{1-(V₁/V₂)^(γ-1)} =
={p₁V₁/(γ-1)}•{1-(V₁/V₂)^(γ-1)} =
=(330000•1.2/0.67) •(1-0.5^0.67)=
=591044.8•0.37 =2.2•10⁵ J

To find the work done by the gas during the adiabatic expansion, we need to use the equation for adiabatic work:

W = (P1V1 - P2V2) / (γ - 1)

Where:
W is the work done by the gas
P1 is the initial pressure of the gas
V1 is the initial volume of the gas
P2 is the final pressure of the gas
V2 is the final volume of the gas
γ is the adiabatic index, which is equal to 5/3 for a monoatomic ideal gas.

Given:
Initial pressure, P1 = 330 kPa
Initial volume, V1 = 1.2 m^3
Final volume, V2 = 2.4 m^3
γ = 5/3

We need to find the final pressure, P2. Since the process is adiabatic, we can use the relationship between pressure and volume for an adiabatic process:

P1V1^γ = P2V2^γ

Plugging in the values:

(330 kPa)(1.2 m^3)^(5/3) = P2(2.4 m^3)^(5/3)

Simplifying:

P2 = (330 kPa)(1.2 m^3)^(5/3) / (2.4 m^3)^(5/3)

Now that we have the final pressure, P2, we can substitute the values into the adiabatic work equation to find the work done by the gas:

W = (P1V1 - P2V2) / (γ - 1)

Plug in the values:

W = (330 kPa)(1.2 m^3) - P2(2.4 m^3) / (5/3 - 1)

Finally, calculate the value for W using a calculator or by simplifying the expression further. This will give you the work done by the gas during the adiabatic expansion.