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Equilibrium Constant and Standard Free Energy Change

The specific ...

Question

The specific heats of argon at constant pressure and constant volume are $525 J / K g$ and $315 J / K g$ , respectively. Its density at NTP will be

1.77 k g / m^{3}

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0.77 k g / m^{3}

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1.77 g / m^{3}

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0.77 g / m^{3}

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Solution

The correct option is A $1.77 k g / m^{3}$
$m (C_{P} - C_{V}) T = P V$
$⟹ C_{P} - C_{V} = \frac{P V}{m T} = \frac{P}{d T}$
$⟹ d = \frac{P}{(C_{P} - C_{V}) T} = \frac{1.01 \times 10^{5}}{273 \times (525 - 315)} = 1.77 k g / m^{3}$

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