A sample of an ideal gas is expanded from $ 1d{m^3} $ to $ 3d{m^3} $ in a reversible process for which $ P = K{V^3} $ , with $ K = \dfrac{1}{5}atm/d{m^3} $ , what is the work done by gas $ (Latm) $ .

Answer
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Hint: Reversible process is defined as the process whose direction can be reversed to return to its original system. For example: extension of springs. When a certain amount of force is applied to the system then it elongates first and when the force is removed after a certain time it attains its original position.

Complete step by step solution:
First of all let us define the reversible system and work done by the system.
Reversible process is defined as the process whose direction can be reversed to return to its original system. For example: extension of springs.
Work done in a reversible system is defined as the product of constant pressure (which is the same for the system and surrounding) and change in volume. Or we can define work done as the $ - \int_{{V_1}}^{{V_2}} {PdV} $ where $ {V_1} $ is the initial volume and $ {V_2} $ is the final volume and $ P $ is a function of volume $ V $ . Here we are given with $ P = K{V^3} $ where $ K = \dfrac{1}{5}atm/d{m^3} $ and $ {V_1} = 1 $ and $ {V_2} = 3 $ . Putting these values in the formula of work done we will get the equation as:
 $ W = - \int_1^3 {\dfrac{1}{5}} {V^3}dV $ and now we know that integration of $ {x^3} $ is $ \dfrac{{{x^4}}}{4} $ hence it will become
 $ W = - \dfrac{1}{5}\mathop {[\dfrac{{{V^4}}}{4}]}\nolimits_1^3 $ .
 $ W = - \dfrac{1}{5} \times [\dfrac{{{3^4} - 1}}{4}] = - \dfrac{1}{5} \times 20 = - 4Latm $ .
Hence, work done by gas $ (Latm) $ where an ideal gas is expanded from $ 1d{m^3} $ to $ 3d{m^3} $ in a reversible process for which $ P = K{V^3} $ , with $ K = \dfrac{1}{5}atm/d{m^3} $ is $ - 4Latm $ .

Note:
Here we are getting the negative sign in the final result which indicates that work is done by the gas and if instead of a negative sign we get a positive sign for the work done then it indicates that work is done on the gas.