First slide

Law of equipartition of energy

Question

The specific heat of argon at constant volume and at constant pressure are 315 JK and 525 JK respectively. Its density at NTP will be

Moderate

A

0.77 g/m³
B

0.77 kg/m³
C

1.77 g/m³
D

1.77 kg/m³

Solution

We know that
$C_{D} - C_{V} = R / M = PV / MT = P / Tρ$
$\begin{array}{l} ∴ ρ = \frac{P}{(C_{p} - C_{V}) T} = \frac{1 \cdot 013 \times 10^{5}}{(525 - 315) \times 273} \\ = \frac{1 \cdot 013 \times 10^{5}}{210 \times 273} = \frac{1 \cdot 013 \times 10^{5}}{57330} \\ = 1 .77 kg / m^{3} \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App