Question

In an isothermal process at 300K, 1 mole of an ideal gas expands from a pressure 100 atm to add against an external pressure of 50atm.Then total entropy change (CalK −1 ) in the process is : A. +0.39 B. -0.39 C. +1.59 D. -1.59

Answer 1

Given that,

Temperature = 300 K

Weight of gas = 1 mol

Initial pressure = 100 atm

Finale pressure = 50 atm

We need to calculate the total entropy change

Using formula of entropy

$\Delta S=nR\ln(\dfrac{P_{1}T_{2}}{P_{2}T_{1}})$

In an isothermal process so the temperature will not change

$\Delta S=nR\ln(\dfrac{P_{1}}{P_{2}})$

Where, n = weight of gas

P₁ =initial pressure

P₂ = final pressure

R = gas constant

Put the value into the formula

$\Delta S=1\times1.9872\ln(\dfrac{100}{50})$

$\Delta S=1.47\ J/K$

The nearest option is 1.59 J/k.

Hence, The total entropy change is 1.59 J/K.

(C) is correct option.