Question

280 mmol of perfect gas occupies 12.7 L at 310 K. Calculate the work done when the gas expands
A) isothermally against a constant external pressure of 0.25 atm
B) isothermally and reversibly
C) into vaccum until its volume has increased by 3.3L.

Answer 1

$n=280\times {10}^{-3}$ models

$V_1=12.7L$ Final volume $\left(v_2\right)=12.7+3.3L=16L$

$T_1=310K$, $P=0.25atm$

A) Gas expands isothermally against a constant external pressure of $0.25atm$.

$\therefore$ Process is irreversible as pressure is constant.

$\therefore$ Work done $=-Pdv$

$=-0.25\left(3.3\right)Latm$

$=-0.25\times 3.3\times 101.33J$

$=-83.6J$

B) Isothermal and reversibly,

Work done $=-nRTln\frac{v_2}{v_1}$

$=-280\times {10}^{-3}\times 8.314\times 310\times ln\frac{16}{12.7}J$

$=-166.7J$

C) Into vaccum, work done$=0J$

($\because$ work done in free expansion process is zero)