Question

Two-litre of $N_2$ at $0^o$C and $5$ atm pressure is expanded isothermally against a constant external pressure of $1$ atm until the pressure of gas reaches $1$ atm, work of expansion is:

[assuming gas to be ideal]

• A. $710.10$ J
• B. $610.10$ J
• C. $810.10$ J
• D. None

Answer 1

The correct option is C $810.10$ J

Using ideal gas equation,

$PV=nRT$

For isothermal condition,

$PV=\text{Constant}$

$\therefore P_1{V}_1=P_2{V}_2.....\left(1\right)$

Given:-

$P_1=5atm$

$P_2=1atm$

$V_1=2L$

$V_2=V\left(\text{say}\right)=?$

Now from ${eq}^n\left(1\right)$, we have

$5\times 2=1\times V\Rightarrow V=10L$

$\therefore V_2=10L$

As we know that, in an irreversible isothermal expansion, the work t=done is given as-

$W=-P_{ext.}\Delta V$

Given that $P_{ext.}=1atm$

$\therefore W=-1(10-2)=-8L-atm=-810.64J\approx -810.10J$

Here $-ve$ sign indicates that the gas is expanding.

Hence the work done of expansion is $810.10J$.

Hence, the correct option is $\text{C}$