A perfectly conducting vessel of volume ` V = 0.4m^3` contains an ideal gas at constant temperature `T = 273 K`. A portion of the gas is let out and the pressure of the falls by `Delta p = 0.24 atm`. (Density of the gas at (STP) is `rho = 1.2 kg//m^3`). Find the mass of the gas which escapes from the vessel.

Explanation

To find the mass of the gas that escapes from the vessel, we can follow these steps:

### Step 1: Use the Ideal Gas Law
The ideal gas law is given by the equation:
\[ PV = nRT \]
where:
- \( P \) = pressure,
- \( V \) = volume,
- \( n \) = number of moles,
- \( R \) = universal gas constant,
- \( T \) = temperature.

### Step 2: Relate Density to Mass and Volume
We know that the density \( \rho \) of the gas is given by:
\[ \rho = \frac{m}{V} \]
where \( m \) is the mass of the gas. Rearranging gives:
\[ m = \rho V \]

### Step 3: Express Number of Moles in Terms of Mass
The number of moles \( n \) can also be expressed as:
\[ n = \frac{m}{M} \]
where \( M \) is the molar mass of the gas.

### Step 4: Substitute into the Ideal Gas Law
Substituting \( n \) into the ideal gas law gives:
\[ PV = \frac{m}{M}RT \]
From this, we can express \( P \) in terms of \( m \):
\[ P = \frac{mRT}{MV} \]

### Step 5: Calculate Initial Pressure
At the beginning, we can calculate the initial pressure using the density:
1. Calculate the initial mass \( m_i \):
\[ m_i = \rho V = 1.2 \, \text{kg/m}^3 \times 0.4 \, \text{m}^3 = 0.48 \, \text{kg} \]
2. Calculate the initial pressure \( P_i \):
\[ P_i = \frac{m_i RT}{V} \]

### Step 6: Calculate Final Pressure
When a portion of the gas is let out, the pressure falls by \( \Delta P = 0.24 \, \text{atm} \). Convert this to Pascals:
\[ \Delta P = 0.24 \times 10^5 \, \text{Pa} = 24000 \, \text{Pa} \]
Thus, the final pressure \( P_f \) is:
\[ P_f = P_i - \Delta P \]

### Step 7: Relate Change in Pressure to Change in Mass
Using the ideal gas law again, we can relate the change in pressure to the change in mass:
\[ \Delta P = \frac{\Delta m \cdot RT}{V \cdot M} \]
Rearranging gives:
\[ \Delta m = \frac{\Delta P \cdot V \cdot M}{RT} \]

### Step 8: Calculate Molar Mass
Using the density and the ideal gas law at STP (Standard Temperature and Pressure):
\[ M = \frac{\rho RT}{P} \]
Substituting the values:
- \( \rho = 1.2 \, \text{kg/m}^3 \)
- \( R = \frac{25}{3} \, \text{J/(mol K)} \)
- \( T = 273 \, \text{K} \)
- \( P = 10^5 \, \text{Pa} \)

Calculating \( M \):
\[ M = \frac{1.2 \times \frac{25}{3} \times 273}{10^5} \]

### Step 9: Substitute Values to Find Mass Escaped
Now substitute \( \Delta P \), \( V \), and \( M \) into the equation for \( \Delta m \):
\[ \Delta m = \frac{24000 \cdot 0.4 \cdot M}{\frac{25}{3} \cdot 273} \]

### Step 10: Calculate the Final Mass Escaped
After performing the calculations, we find:
\[ \Delta m \approx 115.2 \, \text{grams} \]

Thus, the mass of the gas that escapes from the vessel is approximately **115.2 grams**.

---