Question

For a process which follows the equation $P{V^3} = C$, the work done when one mole of ideal gas was taken from 1 atm to $2\sqrt 2$ atm starting from initial temperature of $300K$ is
A.$300R$
B.$150R$
C.$600R$
D.$900R$

Answer 1

Given, the number of moles is equal to one mole and the given gas is an ideal gas. Hence, we can use the ideal gas equation to find out the work done. And the pressure changes from one atm to $2\sqrt 2$atm and given the initial temperature is equal to$300K$. The work done is the amount of energy needed to move the object opposite to the force.

Complete answer:
The value of work done is not equal to $300R$Hence, option (A) is incorrect.
In this process, if applying the given values in the ideal gas equation, the amount of work done will not be equal to $150R$. Hence, the option (B) is incorrect.
According to the question, given
$P{V^3} =$Constant
Here, the work is done in terms of pressure and volume.
Hence, work done $=$ pressure $x$ volume
Therefore we can write,
${P_1}V_1^3 = {P_2}V_2^3$
By rearranging the above equation, will get
$\dfrac{{{P_1}}}{{{P_2}}} = {\left( {\dfrac{{{V_2}}}{{{V_1}}}} \right)^3}$
$3\sqrt {\dfrac{1}{2}} = \dfrac{{{V_2}}}{{{V_1}}}$
By simplifying the equation,
$0.7937 = \dfrac{{{V_2}}}{{{V_1}}}$
Substitute the value of $\dfrac{{{V_2}}}{{{V_1}}}$ in the equation of work done in isothermal process,
$w = - 2.303nRT\log \dfrac{{{V_2}}}{{{V_1}}}$
Where, n is equal to one, R is universal gas constant and T is$300K$. Hence,
$w = - 2.303 \times 1 \times 8.314 \times 300 \times \log 0.7937$
$w = 600K$
Hence, option (C) is correct.
The value of work done will not be equal to $900R$. Hence, the option (D) is correct.

Note:
This process is an isothermal reaction. Hence, in an isothermal reaction, the temperature will not change and it is always constant. And the work done is found out in terms of pressure and volume. Here, $P{V^3} =$is constant. In an isothermal process, the work done is the sum of pressure and volume. And here, we get the work done equal to $600K$.