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प्रश्न
Obtain an expression for the workdone by a gas in an isothermal process.
उत्तर १
Work done by a gas in an isothermal process:
Consider the moles of a gas contained within a cylinder with a moveable, light, and frictionless piston. Let Pi, Vi, and T represent the gas's initial pressure, volume, and absolute temperature, respectively.
Consider an isothermal expansion (or compression) of a gas, where Pf, vf, and T are the final pressure, volume, and absolute temperature of the gas, respectively.
For an isothermal change,
PiVi = PfVf = constant
If the gas behaves like an ideal gas, its equation of state is
PV = nRT = constant ...(i) ....(as T = constant, R is universal gas constant)
The work done in a minuscule isothermal expansion is given by
dW = PdV ...(ii)
The total work done in completing the expansion from initial volume vi to final volume Vf is denoted by
`W = int_(v_i)^(v_f) PdV`
∴ `W = nRT int_(v_i)^(v_f) (dV)/V` ...[from(i)]
∴ W = nRT [In Vf - InVi]
∴ W = nRT In `V_f/V_i`
∴ W = 2.303 nRT `log_10 V_f/V_i`
उत्तर २
Consider the isothermal expansion of an ideal gas. During this process, small work is done, and it is given by
dW = PdV
We get the total work done by integrating the above equation with the limit Vi to Vf.
`W = int_(V_i)^(V_f)Pdv` ...(i)
But we know that for an ideal gas, PV = nRT.
∴ Equation (i) becomes,
`W = int_(V_i)^(V_f)(nRT)/V dV`
`W = nRT int_(V_i)^(V_f)(dV)/V`
`W = nRT [log V]_(V_i)^(V_f)`
`W = nRT log_e (V_f/V_i)`
or `W = 2.303 nRT log_10 (V_f/V_i)`
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