Math, asked by pundyyogesh, 1 year ago

work done during adiabatic reversible expansion (expression)

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Answered by hari98765
2

Analytical Method

Analytically, net work W can be calculated by integrating dW between the limits V1 and V2.

W = \int_{0}^{W}dW = \int_{V_{1}}^{V_{2}}P dV ---- (1)

For an adiabatic change,

PVγ = constant = K (say)

Or, P = K / Vγ

Equation (1) gives,

W = \int_{V_{1}}^{V_{2}}\frac{K}{V^{\gamma }}dV =K\int_{V_{1}}^{V_{2}}V^{-\gamma }dV

Now P1V1γ = P2V2γ =K

So, W = [1/1-γ] [P2 V2γ V2-γ1 - P1 V1γ V11-γ]

Or, W = [1/1-γ] [P2V2 – P1V1] ------- (2)

If T1 and T2 are the initial and final temperatures of the gas,

P1V1 = RT1 and P2V2 = RT2

Substituting in equation (2) we get,

W = [R/1-γ] [T2 - T1]

Or, W = [R/γ-1] [T1 - T2] --------(3)

Therefore, heat required to do the work during expansion is,

H = W/J

= [R / J(γ-1)] [T1 - T2]

As R is gas constant and γ is constant for a given gas, therefore, we find, from equation (3), that work done during an adiabatic expansion is proportional to the fall in temperature. On the other hand, if the gas is compressed, the work is done on the gas and is negative. The work done for adiabatic compression is given by,

W = – [R/γ-1] [T1 - T2]

Or, W = [R/γ-1] [T2 - T1] ------(4)

and

H = [R/γ-1] [T2 - T1]

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