determine the mass of compressed air in a tank. internal volume of the tank is 10L and the pressure of the gas 20.8 MPa. The temperature of the air is 20 C. use the ideal gas law to solve and assume that R equals 287 (J/kg•K)

First, we need to convert the volume from liters to cubic meters, which gives us 0.01 cubic meters. We also need to convert the pressure from megapascals to pascals, which gives us 20.8 million pascals. Finally, we need to convert the temperature from Celsius to Kelvin, which gives us 293 Kelvin.

Now we can rearrange the ideal gas law to solve for the number of moles:

n = (PV) / (RT)

Plugging in the values we have:

n = (20.8e6 Pa x 0.01 m^3) / (287 J/kg*K x 293 K)

n = 0.720 moles

Finally, we can calculate the mass of the compressed air using the number of moles and the molar mass of air, which is approximately 28.97 g/mol:

mass = n x molar mass

mass = 0.720 mol x 28.97 g/mol

mass = 20.9 g

Therefore, the mass of compressed air in the tank is approximately 20.9 grams

msm555 msm555 · Answer 2 · 2024-01-30T10:55:09+01:00

Answer:

[tex] 71.6 [/tex] grams

Explanation:

The Ideal Gas Law relates the pressure, volume, and temperature of an ideal gas. It is given by the equation:

[tex] \Large\boxed{\boxed{\sf PV = nRT}} [/tex]

Where:

[tex] P [/tex] is the pressure of the gas,
[tex] V [/tex] is the volume of the gas,
[tex] n [/tex] is the number of moles of the gas,
[tex] R [/tex] is the ideal gas constant (in this case, given as 287 J/(kg·K)),
[tex] T [/tex] is the temperature of the gas in Kelvin.

To solve for the number of moles ([tex] n [/tex]) of the gas, we need to rearrange the Ideal Gas Law equation:

[tex] \sf n = \dfrac{PV}{RT} [/tex]

Given:

[tex] V = 10 [/tex] L (convert to cubic meters for SI units),
[tex] P = 20.8 [/tex] MPa (convert to Pascals for SI units),
[tex] T = 20 [/tex] °C (convert to Kelvin).

We need to convert the units to SI units:

- [tex] 1 [/tex] L [tex] = 0.001 [/tex] cubic meters,

- [tex] 1 [/tex] MPa [tex] = 10^6 [/tex] Pascals,

- [tex] T [/tex] in Kelvin is [tex] T = 20 \, \textsf{°C} + 273.15 \, \textsf{K} = 293.15 \, \textsf{K} [/tex].

Now, substitute the values into the equation:

[tex] n = \dfrac{(20.8 \times 10^6 \, \textsf{Pa}) \times (10 \times 10^{-3} \, \textsf{m}^3)}{(287 \, \textsf{J per (kg•K)}) \times (293.15 \, \textsf{K})} [/tex]

[tex] n = \dfrac{208 \times 10^5 \,) \times 10^{-4} }{287 \times 293.15 } [/tex]

[tex] n = \dfrac{208 \times 10^{5-4} }{84134.05 } [/tex]

[tex] n = \dfrac{208 \times 10 }{84134.05 } [/tex]

[tex] n = \dfrac{2080}{84134.05} [/tex]

[tex] n \approx \dfrac{2080}{84134.05} [/tex]

[tex] n \approx 2.472245185 [/tex]

So, the number of moles of gas is approximately [tex] 2.472245185 [/tex] moles.

To determine the mass of the gas, we can use the formula:

[tex] \textsf{Mass} = \textsf{Number of moles} \times \textsf{Molar mass} [/tex]

The molar mass of air is approximately [tex] 28.97 [/tex] g/mol.

Therefore,

[tex] \textsf{Mass} = 2.472245185 \, \textsf{moles} \times 28.97 \, \textsf{g/mol} [/tex]

[tex] \textsf{Mass} \approx 71.62094301 \, \textsf{g} [/tex]

[tex] \textsf{Mass} \approx 71.6 \, \textsf{g (in nearest tenth)} [/tex]

So, the mass of the compressed air in the tank is approximately [tex] 71.6 [/tex] grams.