Question

One mole of an ideal gas at $\text{ 300 K }$in thermal contact with surroundings expands isothermally from 1.0 L to 2.0 L against a constant pressure of $\text{ 3}\text{.0 atm }$. In this process, the change in entropy of surrounding $\text{ }\left( \text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{S}}_{\text{surr}}} \right)\text{ }$ in $\text{ J }{{\text{K}}^{-1\text{ }}}$ is $\text{ }\left( 1\text{ L atm = 101}\text{.3 J } \right)$
A) $\text{ 5}\text{.763 }$
B)$\text{ 1}\text{.013 }$
C)$\text{ }-\text{ 1}\text{.013 }$
D) $\text{ }-\text{5}\text{.763 }$

Answer 1

entropy is defined as the change in the heat of the system at the absolute temperature (T).$\text{ }\\!\\!\delta\\!\\!\text{ }{{\text{S}}_{\text{surr}}}\text{ = }\frac{\text{ }\\!\\!\delta\\!\\!\text{ }{{\text{q}}_{\text{surr}}}}{\text{T}}\text{ }$ .during the expansion the heat is gained by the surrounding thus from the first law of thermodynamics the $\text{ }\\!\\!\delta\\!\\!\text{ q = }\\!\\!\delta\\!\\!\text{ U + }\\!\\!\delta\\!\\!\text{ w }$ change in the heat will be equal to the$-\text{ }\\!\\!\delta\\!\\!\text{ }{{\text{w}}_{\text{Surr}}}$.

Complete step by step solution:
We are given the following data:
The number of moles of gas is, $\text{ n = 1 }$
The absolute temperature of an ideal gas is equal to, $\text{ 300 K }$
The gas expands isothermally from the volume $\text{ }{{\text{V}}_{\text{1}}}=\text{ 1}\text{.0 L }$ to the volume $\text{ }{{\text{V}}_{2}}=\text{ 2}\text{.0 L }$ against the constant pressure of $\text{ P = 3}\text{.0 atm }$
We are interested to determine the change in the entropy of the system $\text{ }\Delta \text{S }$
The entropy is a measure of the randomness of the system. Mathematically it is expressed as the ratio of change in heat per unit absolute temperature .the relation is stated as follows,
$\text{ }\\!\\!\delta\\!\\!\text{ }{{\text{S}}_{\text{surr}}}\text{ = }\frac{\text{ }\\!\\!\delta\\!\\!\text{ }{{\text{q}}_{\text{surr}}}}{\text{T}}\text{ }$
Where $\text{ }\\!\\!\delta\\!\\!\text{ }{{\text{q}}_{\text{surr}}}\,\text{ }$ is the heat gained by the surrounding and T is the absolute temperature.
If U is the internal enthalpy of a system. Suppose the system absorbs the heat q from the surrounding and also performs some work which is equal to w. The relation between the internal enthalpy, heat, and work is given as follows,
$\text{ }\\!\\!\delta\\!\\!\text{ q = }\\!\\!\delta\\!\\!\text{ U + }\\!\\!\delta\\!\\!\text{ w }$ (1)
The isothermal process is a thermodynamic process in which the temperature of the system remains constant. That is $\text{ }\Delta \text{T = 0 }$ . Since for ideal gas, the internal enthalpy U depends on the temperature, for an isothermal expansion process the internal enthalpy is equal to zero,$\text{ }\delta \text{U = 0 }$ .
Thus equation (1) becomes,
$\text{ }\\!\\!\delta\\!\\!\text{ q = }\\!\\!\delta\\!\\!\text{ w }$ (2)
Now, heat is gained by the surrounding and heat is lost by the system. Therefore, we can write the relation (2) as,
$\text{ }\\!\\!\delta\\!\\!\text{ }{{\text{q}}_{\text{system}}}\text{ = }-\text{ }\\!\\!\delta\\!\\!\text{ }{{\text{w}}_{\text{Surr}}}\text{ }$ (3)
The entropy of the surrounding is given by the ratio of the heat gained by the surrounding to the absolute temperature. Thus the entropy for surrounding written as,
$\text{ }\\!\\!\delta\\!\\!\text{ }{{\text{S}}_{\text{surr}}}\text{ = }\frac{\text{ }\\!\\!\delta\\!\\!\text{ }{{\text{q}}_{\text{surr}}}}{\text{T}}\text{ }$
From (3), the entropy is written as,
$\text{ }\\!\\!\delta\\!\\!\text{ }{{\text{S}}_{\text{surr}}}\text{ = }\frac{\text{ }\\!\\!\delta\\!\\!\text{ }{{\text{q}}_{\text{surr}}}}{\text{T}}\text{ = }\frac{-\text{ }\\!\\!\delta\\!\\!\text{ w}}{\text{T}}\text{ }$
The work done is written as the product of pressure and change in volume.
$\text{ }\\!\\!\delta\\!\\!\text{ }{{\text{S}}_{\text{surr}}}\text{ = }\frac{-\text{P}\left( \text{dV} \right)}{\text{T}}\text{ = }\frac{-\text{P}\left( {{\text{V}}_{\text{2}}}-{{\text{V}}_{\text{1}}} \right)}{\text{T}}\text{ }$
Substitute the values in the equation .we have,
$\begin{aligned} & \text{ }\\!\\!\delta\\!\\!\text{ }{{\text{S}}_{\text{surr}}}\text{ = }\frac{-3\left( 2.0-1.0 \right)}{300}\text{ }\times \text{ 103 Joules } \\\ & \therefore \text{ }\\!\\!\delta\\!\\!\text{ }{{\text{S}}_{\text{surr}}}\text{ = }-1.013\text{ joules } \\\ \end{aligned}$
The change in entropy for the ideal gas is equal to $\text{ }-1.013\text{ }$ joules.

Hence, (C) is the correct option.

Note: Note that, if the work is done by the surrounding on the system (i.e. compression of gas) then the w is taken as the positive, and for isothermal process the relation becomes$\text{ }\\!\\!\delta\\!\\!\text{ q = }-\text{ }\\!\\!\delta\\!\\!\text{ w }$. However, when work is done by the system on the surrounding (i.e. expansion) the w took as the negative, and for isothermal process the relation becomes$\text{ }\\!\\!\delta\\!\\!\text{ q = }\\!\\!\delta\\!\\!\text{ w }$.