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2. An LPG tank contains 11.0 kg of butane gas, (C4H10), in a 25.0 L gas tank. what must be the pressure of the gas inside the tank when the temperature is 25°C?​

Sagot :

✏️IDEAL GAS EQUATION

[tex]\tt{\huge{\green{Solution:}}}[/tex]

The problem asks to solve for the pressure of the gas inside the tank. First, we need to identify the quantities to be used. These are

P = pressure

V = volume

n = number of moles

R = universal gas constant = 0.082057 L • atm/K • mol

T = absolute temperature

m = mass

MW = molecular weight

Second, we need to identify the given values. These are

V = 25.0 L

T = 25°C + 273.15 = 298.15 K

m = 11.0 kg = 1.10 × 10⁴ g

Third, we need to find the molecular weight of gas. Since the given gas is butane, the molecular weight is

MW = (12.01 g/mol × 4) + (1.008 g/mol × 10)

MW = 58.12 g/mol

Fourth, we need to find the number of moles of gas. The number of moles of butane gas is

[tex]n = \dfrac{m}{\text{MW}}[/tex]

[tex]n = \dfrac{1.10 \times 10^{4} \: \text{g}}{\text{58.12 g/mol}}[/tex]

n = 189.26 mol

Finally, we can now solve for the pressure of the gas. Therefore, the pressure of the gas is

[tex]P = \dfrac{nRT}{V}[/tex]

[tex]P = \dfrac{(\text{189.26 mol})(0.082057 \: \text{L} \: \cdot \: \text{atm/}\text{K} \: \cdot \: \text{mol})(\text{298.15 K})}{\text{25.0 L}}[/tex]

[tex]\boxed{P = \text{185 atm}}[/tex]

[tex]\\[/tex]

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