Question

50g of oxygen at NTP is compressed adiabatically to a pressure of 5 atmosphere. The work done on the gas, if $\gamma$ = 1.4 and R = 8.31 J mol–1K–1 is

• A. –5173 J
• B. 1131 J
• C. –1364 J
• D. 5673 J

Answer 1

Answer: (A)

–5173 J

Answer 2

: Here, $T _ { 1 } = 273 \mathrm {~K} , P _ { 1 } = 1 \mathrm {~atm}$

$P _ { 2 } = 5 \mathrm {~atm}$

No. of moles of oxygen

$= \frac { 50 } { 32 }$

$\Rightarrow T _ { 2 } = T _ { 1 } \left( \frac { P _ { 1 } } { P _ { 2 } } \right) ^ { \frac { \gamma - 1 } { \gamma } } = 273 \left( \frac { 5 } { 1 } \right) ^ { \frac { 14 - 1 } { 1.4 } }$

$= 273 \times ( 5 ) ^ { 2 / 7 } = 273 \times 1.584$ $= 432.37$

$W = \frac { n R } { \gamma - 1 } \left[ T _ { 1 } - T _ { 2 } \right]$ $= \frac { 50 } { 32 } \times \frac { 8.31 } { 1.4 - 1 } ( 273 - 432.37 )$

$= 5173 \mathrm {~J}$