Two moles of an ideal monoatomic gas at NTP are compressed adiabatically and reversibly to occupy a volume of $4.48 {d m}^{3}$ . Calculate the amount of work done, $Δ U$ , final temperature and pressure of the gas. $C_{V}$ for ideal gas $= 12.45 J K^{- 1}$ ${m o l}^{- 1}$

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Solution

For an ideal gas, $γ = \frac{C_{P}}{C_{V}} = 1.667$
Initial volume, $V_{1} = 2 \times 22.4 = 448 {d m}^{3}$
Initial pressure, $P_{1} = 1 a t m$
Initial temperature, $T_{1} = 273 K$
Final volume, $V_{2} = 4.48 {d m}^{3}$
Let the final pressure be $P_{2}$ and temperature be $T_{2}$ .
Applying $P_{1} {V_{1}}^{γ} = P_{2} {V_{2}}^{γ}$
$\frac{P_{1}}{P_{2}} = {(\frac{V_{2}}{V_{1}})}^{γ} = {(\frac{4.48}{44.8})}^{1.667}$
or $\frac{P_{2}}{P_{1}} = {(10)}^{1.667}$
$P_{2} = {(10)}^{1.667} (P_{1} = 1$ given)
$log P_{2} = 1.667 log 10 = 1.667$
$P_{2} = a n t i l o g 1.667 = 46.45 a t m$
Final temperature $= \frac{P_{2} V_{2}}{P_{1} V_{1}} . T_{1} = \frac{46.45 \times 4.48}{1 \times 44.8} \times 273 = 1268 K$
Work done on the system $= n . C_{V} . Δ T$
$= 2 \times 12.45 \times (1268 - 273)$
$= 2 \times 12.45 \times 995 = 24775.5 J$
From the first law of thermodynamics,
$Δ U = q + w = 0 + 24775.5 = 24775.5 J$

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