Question

The equation of state for a gas is given by $PV = nRT + \alpha V$, where $n$ is the number of moles and $\alpha$ is a positive constant. The initial temperature and pressure of one mole of the gas contained in a cylinder are ${T_0}$ and ${P_0}$ respectively. The work done by the gas when its temperature doubles isobarically will be:
A) $\dfrac{{{P_0}{T_0}R}}{{{P_0} - \alpha }}$
B) $\dfrac{{{P_0}{T_0}R}}{{{P_0} + R}}$
C) ${P_0}{T_0}R\ln 2$
D) ${P_0}{T_0}R$

Answer 1

Work done in a finite process, for which the volume changes from ${V_1}$ to ${V_2}$ is W = \int {dW = \mathop \smallint \nolimits_{{V_1}}^{{V_2}} } pdV. Also, for one mole gas $n = 1$.

Complete step by step solution: A closed thermodynamic system, i.e., a system having a fixed mass, can be described completely by three of its properties – volume $\left( V \right)$, pressure $\left( p \right)$, and temperature $\left( T \right)$. All other properties of the system depend on and therefore are functions of $V,p$, and $T$. In the isobaric process, pressure remains constant, and volume and the temperature change their values.
Work done in a finite process, for which the volume changes from ${V_1}$ to ${V_2}$ is
W = \int {dW = \mathop \smallint \nolimits_{{V_1}}^{{V_2}} } pdV
We know that for one mole gas $n = 1$ and it is given that the initial temperature and pressure of one mole of the gas contained in a cylinder are ${T_0}$ and ${P_0}$ respectively.
Let, ${V_i}$ is the initial volume. So, initially, the equation is ${P_0}{V_i} = R{T_0} + \alpha {V_i}$ $\therefore {V_i} = \dfrac{{R{T_0}}}{{{P_0} - \alpha }}$
Let, ${V_f}$ is the final volume. So, at the final stage, the equation is ${P_0}{V_f} = R.2{T_0} + \alpha {V_f}$ $\therefore {V_f} = \dfrac{{2R{T_0}}}{{{P_0} - \alpha }}$
The work done will be W = \mathop \smallint \nolimits_{{V_i}}^{{V_f}} {P_0}dV = {P_0}\left[ V \right]_{{V_i}}^{{V_f}} = {P_0}\left( {{V_f} - {V_i}} \right) = {P_0}\left( {\dfrac{{2R{T_0}}}{{{P_0} - \alpha }} - \dfrac{{R{T_0}}}{{{P_0} - \alpha }}} \right) = \dfrac{{{P_0}{T_0}R}}{{{P_0} - \alpha }}

Additional information: From the expression of work done, we can know that when there is no change in volume, and $dV = 0$. So, $dW = 0$. So, no work is done in such a process. Also, when the volume increases, i.e., for expansion, $dV$ is positive, i.e., $dV > 0$. So, $dW > 0$; Positive amount of work is done in this process. The system releases some energy to its surroundings; it is termed as work done by the system. Again, when the volume decreases, i.e., when $dV$ is negative, i.e., $dV < 0$. So, $dW < 0$; a negative work is done in this process. This system receives some energy from its surroundings; it is termed as work done on the system.

Note: When the pressure of a system changes in a process, the integral cannot be evaluated unless pressure can be expressed as a function of volume. It is only possible if the equation of state of the system is known.