Question

One mole of $O_2$ gas having a volume equal to $22.4$ Litres at $0^{\circ}C$ and $1$ atmospheric pressure in compressed isothermally so that its volume reduces to $11.2$ litres. The work done in this process is -

• A. 1672.5 J
• B. 1728 J
• C. - 1728 J
• D. - 1572.5 J

Answer 1

Answer: (D)

Work done in adiabatic process,
$ W =\mu RT \log _{ e } \frac{ V _{ e }}{ V _{ r }} $
$W =1 \times 8.314 \times 273 \times \log _{ e }\left(\frac{11.2}{22.4}\right) $
$ W =8.314 \times 273 \times \log _{ e }\left(\frac{1}{2}\right) $
$ W =8.314 \times 273 \times\left[\log _{ e }(1)-\log _{ e }(2)\right] $
$ W =8.314 \times 273 \times(0-0.69)$
$W =-1566.1 \,J$