Question

Calculate the work done by one mole of a van der waals gas if during its isothermal expansion its volume increases from 1m^3 to 2m^3 at a temperature 300k.take a=1.39×10^-6 atm m^6 mol^-2 and b= 39.1×10^-6m^3 mol^-1

Answer 1

Answer on Question #40847, Physics, Molecular Physics

Calculate the work done by one mole of a van der Waals gas if during its isothermal expansion its volume increases from $1\mathrm{m}^3$ to $2\mathrm{m}^3$ at a temperature 300 K. Take $a = 1.39\times 10^{-6}$ atm $\mathrm{m}^6\mathrm{mol}^{-2}$ and $b = 39.1\times 10^{-6}\mathrm{m}^3\mathrm{mol}^{-1}$.

Solution:

We can start from the definition of work,

$$W = - \int P d V$$

An expression for $P$ is required – this comes from the van der Waals equation. Solving the van der Waals equation for $P$ leads to

$$P = \frac {n R T}{V - n b} - \frac {a n ^ {2}}{V ^ {2}}$$

The equation uses the following state variables: the pressure $P$, total volume $V$, number of moles $n$, and absolute temperature of the system $T$.

Substituting this relation into the equation for work yields

$$W = - \int_ {V _ {1}} ^ {V _ {2}} \left(\frac {n R T}{V - n b} - \frac {a n ^ {2}}{V ^ {2}}\right) d V$$

Since the process is isothermal, $T$ can be pulled out of the integral along with the other constants to give

$$W = - n R T \int_ {V _ {1}} ^ {V _ {2}} \frac {1}{V - n b} d V + a n ^ {2} \int_ {V _ {1}} ^ {V _ {2}} \frac {1}{V ^ {2}} d V$$

Integrating leads to the result

$$W = - n R T \ln \left(\frac {V _ {2} - n b}{V _ {1} - n b}\right) - a n ^ {2} \left(\frac {1}{V _ {2}} - \frac {1}{V _ {1}}\right)$$

1 atm = 101325 Pa

$a = 1.39 \times 10^{-6}$ atm $\mathrm{m}^6\mathrm{mol}^{-2} = 1.39 \cdot 101325 \cdot 10^{-6} = 0.1408\mathrm{Pa} \cdot \mathrm{m}^6 \cdot \mathrm{mol}^{-2}$

So,

$$W = - 8.31 \cdot 300 \cdot \ln \left(\frac {2 - 39.1 \cdot 10 ^ {-6}}{1 - 39.1 \cdot 10 ^ {-6}}\right) - 0.1408 \cdot \left(\frac {1}{2} - \frac {1}{1}\right) = - 1727.99$$

Answer. $W = -1728$ J.