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Work

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Question

Find the work requires to compress adiabatically $1 g$ of air initially at $N T P$ to half its volume. Density of air at $N T P = 0.001129 g c m^{- 3}$ and $\frac{C_{p}}{C_{v}} = 1.4$

62.64 J

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32.64 J

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- 32.64 J

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- 62.64 J

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Solution

The correct option is D $- 62.64 J$
For an adiabatic process
$T_{1} V_{1}^{γ - 1} = T_{2} V_{2}^{γ - 1}$
or $T_{2} = T_{1} {(\frac{V_{1}}{V_{2}})}^{γ - 1} = 273 (2)^{0.4} = 360 K$
$V = \frac{1 \times 10^{- 6}}{0.00129} m^{3}$

We find $R$ for $1 g$ of air using $P V = R T$
$R = \frac{76 \times 13.6 \times 981 \times 10^{- 6}}{273 \times 0.00129} = 2.88$
W = $\frac{R}{γ - 1} (T_{1} - T_{2}) = \frac{2.88}{0.4} (273 - 360)$
$W = - 62.64 J$

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Similar questions

(\frac{C_{P}}{C_{V}} f o r a i r = 1.4)

1.293 k g m^{- 3}

0^{\circ} C

13.6 \times 10^{3} k g m^{- 3}

C_{p} = 0.2417 c a l k g^{- 10} C^{- 1}

C_{v} = 0.1715

100^{\circ} C

(g = 9.8 N k g^{- 1})

\frac{C_{p}}{C_{v}}

3^{0.4} = 1.55, 3^{1.4} = 4.66

N T P

1.3 g/L

γ = 1.4

27^{\circ} C

3

γ = 1.4

3^{1.4} = 4.655

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