Question: If a gas expands with temperature according to the relation \[V = K{T^{2/3}}\], then what will be wo...

Question

If a gas expands with temperature according to the relation $V = K{T^{2/3}}$ , then what will be work done by the gas when the temperature changes by $30^\circ C$ ?
A. $40R$
B. $30R$
C. $20R$
D. $10R$

Answer 1

Solution

Expansion or compression against constant external pressure is the way in which gases do work. This work done by gases is also called the Pressure-Volume work (PV work). The energy of the system can change due to work and other forms of energy transfer such as heat. To solve this problem we use the work done formula and using the relation given in the question we will get the required work done.

Complete step by step solution:
We are going to use the formula for the work done as given below,
$W = \int {PdV}$ …… (1)
Here $P$ is the pressure
$dV$ is the change in the volume
We know from the perfect gas equation
$PV = RT$
Rearranging the above equation to get the pressure,
$P = \dfrac{{RT}}{V}$ …… (2)
Substituting the equation (2) in (1)
$W = \int {\dfrac{{RT}}{V}dV}$ ……. (3)
Since it is given in the question that the volume changes according to the relation,
$V = K{T^{2/3}}$ …….. (4)
Differentiating equation (4)
$dV = \dfrac{2}{3}K{T^{ - 1/3}}dT$ ……. (5)
Dividing the change in the volume by volume we get,
$\dfrac{{dV}}{V} = \dfrac{2}{3}\dfrac{{dT}}{T}$ …… (6)
Substituting the equation (6) in (3)
$W = \int_{{T_1}}^{{T_2}} {\dfrac{2}{3}} \dfrac{{RT}}{T}dT$
$W = \dfrac{2}{3}R\int_{{T_1}}^{{T_2}} {dT}$
$W = \dfrac{2}{3}R({T_2} - {T_1})$
Given that the change in the temperature is ${T_2} - {T_1} = 30^\circ C$ . Substituting in the above equation,
$W = \dfrac{2}{3}R(30^\circ C)$
$W = 20R$
Therefore the correct option is C.

Note:
The pressure-volume work is only valid as long as the process is quasistatic. Quasistatic is an infinitesimally slower process. If the process is not in the quasi static state then the work done formula will be changed into $dW = PdV + dW(other)$ . Other works include shaft work. This quasi-static state is the condition of a reversible process. But no process in the real world is reversible.