The problem states that there are 4 kg of air in a bathyscaphe at a temperature of 20.6 °C and a pressure of $1 \times 10^5$ Pa. A person inside uses 778 cm³ of air per minute. The molar mass of air is given as $M_0 = 29 \times 10^{-3}$ kg/mol. We need to calculate how many hours a person can stay in the bathyscaphe.

1. Problem Description

The problem states that there are 4 kg of air in a bathyscaphe at a temperature of 20.6 °C and a pressure of $1 \times 10^5$ Pa. A person inside uses 778 cm³ of air per minute. The molar mass of air is given as $M_0 = 29 \times 10^{-3}$ kg/mol. We need to calculate how many hours a person can stay in the bathyscaphe.

2. Solution Steps

First, we need to find the volume of the air in the bathyscaphe using the ideal gas law. The ideal gas law is given by:

$PV = nRT$

where:

$P$ is the pressure,

$V$ is the volume,

$n$ is the number of moles,

$R$ is the ideal gas constant ($8.314$ J/(mol*K)),

$T$ is the temperature in Kelvin.

We are given the mass of the air, $m = 4$ kg, and the molar mass $M_0 = 29 \times 10^{-3}$ kg/mol. We can find the number of moles:

$n = \frac{m}{M_0} = \frac{4}{29 \times 10^{-3}} = \frac{4000}{29}$ moles.

Convert the temperature from Celsius to Kelvin:

$T = 20.6 + 273.15 = 293.75$ K.

Now we can find the volume:

$V = \frac{nRT}{P} = \frac{(\frac{4000}{29}) \times 8.314 \times 293.75}{1 \times 10^5} = \frac{4000 \times 8.314 \times 293.75}{29 \times 10^5} = \frac{9742700}{2900000} \approx 3.35955$ m³.

The person uses 778 cm³ of air per minute. Convert this to m³:

$778 \text{ cm}^3 = 778 \times 10^{-6} \text{ m}^3 = 0.000778 \text{ m}^3$.

Now, we can calculate how many minutes the air will last:

$\text{Time in minutes} = \frac{\text{Total Volume}}{\text{Consumption per minute}} = \frac{3.35955}{0.000778} \approx 4318.19$ minutes.

Finally, convert minutes to hours:

$\text{Time in hours} = \frac{4318.19}{60} \approx 71.97$ hours.

3. Final Answer

The person can stay in the bathyscaphe for approximately 71.97 hours.