From given data, calculate the value of the mechanical equivalent of heat. The specific heat capacity of air at constant volume is 170 cal/kg-K, $γ = C_{P} / C_{V} = 1.4$ and the density of air at STP is $1.29 k g / m^{3}$ . Gas constant $R = 8.3 J / m o l K$ .

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Solution

Using $p V = n R T$ , the volume of 1 mole of air at STP is
$V = \frac{n R T}{p} = \frac{(1 m o l) \times (8.3 J / m o l - K) \times (273 K)}{1.01 \times 10^{5} N / m^{2}} = 0.0224 m^{3}$

The mass of 1 mole is, therefore,
$(1.29 k g / m^{3}) \times (0.0224 m^{3}) = 0.029 k g$ .
The number of moles in 1 kg is $\frac{1}{0.029}$ . The molar heat capacity at constant volume is
$C_{v} = \frac{170 c a l}{(1 / 0.029) m o l - K} = 4.93 c a l / m o l - K$ .
Hence, $C_{p} = γ C_{v} = 1.4 \times 4.93 c a l / m o l - K$
or, $C_{p} - C_{v} = 0.4 \times 4.93 c a l / m o l - K$
$= 1.97 c a l / m o l - K$
Also, $C_{p} - C_{v} = R = 8.3 J / m o l - K$ .
Thus, $8.3 J = 1.97 c a l$ .
The mechanical equivalent of heat is
$\frac{8.3 J}{1.97 c a l} = 4.2 J / c a l$ .

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