Question

One mole of oxygen is allowed to expand isothermally and reversibly from $5{m^3}$ to $10{m^3}$ at $300k$ . Calculate the work done in expansion of the gas.

Answer 1

Since, we know that the work done in the isothermal irreversible process is given by the equation, $w = - 2.303nRTln\left( {\dfrac{{{V_f}}}{{{V_i}}}} \right)$ . And number of moles, absolute temperature of the gas, final volume and initial volume of the gas is given in the question. Thus, substituting the values in the equation, we can calculate the work done in expansion of the gas.

Complete step by step answer:
The work done in the isothermal irreversible process is given by the equation,
$w = - 2.303nRTln\left( {\dfrac{{{V_f}}}{{{V_i}}}} \right)$
Where,
n = number of moles
R = universal gas constant
T = absolute temperature of the gas
${V_f} =$ final volume
${V_i} =$ initial volume
Now given in the question are,
n = 1 mol
${V_f} =$ 10 m3
${V_i} =$ 5 m3
T = 300 K
So, substituting the values in the equation, $w = - 2.303nRTln\left( {\dfrac{{{V_f}}}{{{V_i}}}} \right)$ , we get
$w = - \left( {2.303} \right)\left( {1{\text{ }}mol} \right)\left( {8.314{\text{ }}\dfrac{J}{{Kmol}}} \right)\left( {300{\text{ }}K} \right)ln\left( {\dfrac{{10{m^3}}}{{5{m^3}}}} \right)$
We know that, $R = 8.314\dfrac{J}{{Kmol}}$
$w = - 1728.98{\text{ }}J$

Thus, the work done in expansion of the gas is $- 1728.98{\text{ }}J$

Note: The negative sign indicates that the work is done by the system on the surroundings. An isothermal process is a process which is conducted in a manner such that the temperature remains constant during the entire operation. It may be easier in most cases to the work through the heat added, but if you only have information about the pressure, volume or temperature, one of these equations could simplify the problem. Since work is a form of energy, its unit is the joule (J).