Question

The q value and work done in isothermal reversible expansion of one mole of an ideal gas from an initial pressure of 10 bar to final pressure of 1 bar at constant temperature 273K are:
a.) 5.22 KJ, -5.22 KJ
b.) -5.22 KJ, 5.22 KJ
c.) 5.22 KJ, 5.22 KJ
d.) -5.22 KJ, -5.22 KJ

Answer 1

Hint: Try to recall that in an isothermal process, temperature remains constant throughout the process or $\Delta T = 0$ and also work done in an isothermal reversible expansion of an ideal gas is given by, $W = - 2.303nRT{\log _{10}}\dfrac{{{P_1}}}{{{P_2}}}$ where, n = number of mole of an ideal gas, P1=initial pressure of the gas, P2=final pressure of the gas, R = universal gas constant and its value is 8.3124J/mol/K and T is the temperature at which process takes place. Now by using this formula you can easily answer the given question.

Complete step by step solution: It is known that you can easily find the work done in case of isothermal reversible expansion of an ideal gas by using the formula, $W = - 2.303nRT{\log _{10}}\dfrac{{{P_1}}}{{{P_2}}}$.
Calculations:
Given, P1= 10 bar
P2= 1 bar, n=1 and T= 273 K and R= 8.314 J/mol/K.
Now, $W = - 2.303nRT{\log _{10}}\dfrac{{{P_1}}}{{{P_2}}}$
$\Rightarrow W = - 2.303(1)(8.314)(273){\log _{10}}\dfrac{{10}}{1}$
$\Rightarrow W = ( - 5227.16)({\log _{10}}10)$
$\Rightarrow W = - 5227.16J$ $\left( {\because {{\log }_{10}}10 = 1} \right)$
$or,W = - 5.22KJ$
Now, since it is known to you that the process is isothermal. So, $\Delta T = 0$
It is known that, $\Delta U = \dfrac{f}{2}R\Delta T$
$\Rightarrow \Delta U = \dfrac{f}{2}R(0)$
$or,\Delta U = 0$.
Since, W and $\Delta U$ is now known , so you can easily find q from first law of thermodynamics: $\Delta U = q + W$
$\Delta U = 0$ and $W = - 5.22KJ$.
From first law of thermodynamics,
$\Delta U = q + W$
$\Rightarrow 0 = q + ( - 5.22)$
$or,q = 5.22KJ$
Therefore, from above calculations we can easily conclude that option A is the correct option to the given question.

Note: It should be remembered to you that the internal energy of gas depends only on its temperature and does not depend on the amount of gas.
Also, you should remember that internal energy is a state function whereas work done and heat are path functions