Question

A large tank is filled with methane gas at a concentration of 0.740 kg/m^3. The valve of a 1.90-m pipe connecting the tank to the atmosphere is inadvertantly left open for 10.0 hours. During this time, 6.10×10^-4 kg of methane diffuses out of the tank, leaving the concentration of methane in the tank essentially unchanged. The diffusion constant for methane in air is 2.10x10^-5 m/. What is the cross-sectional area of the pipe? Assume that the concentration of methane in the atmosphere is zero.

Answer 1

To calculate the cross-sectional area of the pipe, we can use the following equation:

```

A = m / (C * D * t)

```

where:

* A is the cross-sectional area of the pipe (m^2)

* m is the mass of methane that diffuses out of the tank (kg)

* C is the concentration of methane in the tank (kg/m^3)

* D is the diffusion constant for methane in air (m^2/s)

* t is the time that the valve is left open (s)

We are given the following values:

* m = 6.10×10^-4 kg

* C = 0.740 kg/m^3

* D = 2.10x10^-5 m^2/s

* t = 10.0 * 3600 s

Substituting these values into the equation, we get:

```

A = 6.10×10^-4 kg / (0.740 kg/m^3 * 2.10x10^-5 m^2/s * 10.0 * 3600 s)

```

```

A = 0.0010903760903760902 m^2

```

Therefore, the cross-sectional area of the pipe is **0.0010903760903760902 m^2**.

Step-by-step explanation:

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