Answer to Question #88588 in Molecular Physics | Thermodynamics for Sridhar

Question

An open flask contains air at 27°C. To what temperature at which 2/5 th mass of air escapes the flask

Answer 1

According to Charle's law,

$$\frac {V_1} {T_1}=\frac {V_2} {T_2} (1)$$

Using (1) we got:

$$T_2=\frac {T_1 \times V_2} {V_1} (2)$$

In our case,

T₁=300 K,

$$V_1=\frac {3V_2} {5}$$

We got:

T₂=500 K or 227 °C

Answer:

500 K or 227 °C