Question

Calculate the maximum work done when pressure on 10 g of hydrogen is reduced from 20 to 1 atm at constant temperature of 273 K. The gas behaves ideally. Also calculate Q. $\left( {R = 2{\text{ }}cal/K{\text{ }}mol} \right)$

Answer 1

Workdone is pressure volume work during compression or expansion of gas. Pressure is decreasing in this question and q is the amount of heat change in this process.

Step by step solution:
Mass of hydrogen is given as 10 g.
The no. of moles of hydrogen can be found as:
$moles = \dfrac{{mass}}{{Molecular\;mass}}$
Molecular mass of hydrogen $\left( {{H_2}} \right)$ is 2.
${n_{{H_2}}} = \dfrac{{10}}{2}$
${n_{{H_2}}} = 5$moles
As given at constant temperature $T = 273\;K$(means isothermal process)
Initial pressure ${P_i}$ is 20 atm
Final pressure ${P_f}$ is 1 atm
As we know, the work done in an isothermal reversible process is:
$W = 2.303nRT\log \left( {\dfrac{{{P_i}}}{{{P_f}}}} \right)$
Where n is number of moles and R is universal gas constant.
The value of R is $8.314\dfrac{{{m^3}.Pa}}{{mol.K}}$
Now on substituting the values of$n$, $T$, ${P_i}$and ${P_f}$
$W = 2.303 \times 5 \times 2 \times 273 \times \log \left( {\dfrac{{20}}{1}} \right)$
$W = 8179.82{\text{ cal}}$
Now as we know
$1{\text{ }}kcal = 1000{\text{ }}cal$
So,
$W = 8.18{\text{ kcal}}$
Next to find Q:
As we know for ideal gas at constant temperature, change in internal energy will be zero i.e. $\Delta U = 0$
From first law of thermodynamics: The change in internal energy $\left( {\Delta U} \right)$ of a system can be expressed as the difference of heat transfer $\left( {\Delta Q} \right)$ into a system and the work done by the system $\left( W \right)$
In mathematical expression, it can be written as:
$\Delta U = Q - W$
On substituting the values
$0 = Q - (8.18)$
$Q = 8.18\;kcal$
Hence the maximum work done by the system is 8.18 kcal and the heat added to the system is 8.18 kcal.

Note: In chemistry, the convention is that anything going out of the system is negative and anything coming into the system is positive.
$Q$ is positive if heat is added to the system, and negative if heat is removed.
$W$ is positive if work is done by the system, and negative if work is done on the system.