the mass of butane gas in the gas cylinder

Question

Universal Gas Constant (R) : 0.08206 atm-L/mol-K or 8.314 J/mol-K

Consider a gas cylinder with volume of 31200dm³ for storing and shipping pure butane gas (C₄H₁₀), it is known that the pressure in the bottle is 41kPa at 15^oC.

(a) Determine the mass of butane gas in the gas cylinder.

(b) Given that the highest pressure that the gas cylinder can tolerate is 55.2kPa or leakage will be resulted, determine whether the gas cylinder with conditions listed above can be stored under 45^oC environment.

Answer 1

ASSUMPTION: dealing with an ideal gas

This will require the use of the Ideal Gas Law: pV = nRT

STEP 1:

convert the provided state-values to units that match those of the Universal Gas Constant

R [=] (atm·L)/(mol·K)

1. Convert Volume: dm³ → cm³ → mL → L

_{note: if 1 dm is 0.1m and 1cm is 0.01m, then there are 10cm for every 1dm}

_{note: (a/b)3 = a3/b3}

_{note: 1cm3 = 1cc(cubic centimeter) = 1mL}

31,200dm³ x (10cm/1dm)³ x (1mL/1cm³) x (1L/1000mL) = 31,200 L

_{note: it turns out 1dm3 = 1L, how convenient!}

1. Convert Pressure: kPa → atm

41kPa x (1atm/101.325kPa) = 0.40 atm

1. Convert Temperature: °C → K

15 °C + 273.15 = 288.15 K

A)

STEP 2:

use Algebra to solve the Ideal Gas Law for n (moles), then evaluate the number of moles of Butane

n = pV/(RT)

= (0.40atm · 31,200L) / (0.08206 (atm·L/mol·K) · 288.15K)

= 528 mol C₄H₁₀ (Butane)

_{side note: analyze the units to confirm}

^{[=] atm/1 x L/1 x 1/(atm·L)/(mol·K) x 1/K}

^{atm/1 x L/1 x (mol·K)/(atm·L) x 1/K}

^{atm/1 x L/1 x (mol·K)/(atm·L) x 1/K}

^mol √

STEP 3:

use dimensional analysis to convert moles of C₄H₁₀ to grams of C₄H₁₀

_{note: 1 mol C4H10 = 4(12.011g C) + 10(1.008g H) = 58.124g C4H10}

528 mol C₄H₁₀ x (58.124g C₄H₁₀ / 1mol C₄H₁₀) ~= 31,000g C₄H₁₀

ANSWER:

31,000g C₄H₁₀ gas in the cylinder

B)

given that p_max = 55.2kPa (.545 atm), determine whether p(T=45°C) < p_max

_{note: the p_max of the gas tank is less than the pressure of Earth's atmosphere}

STEP 4:

use Algebra to solve the Ideal Gas Law for p (pressure), then evaluate the pressure of Butane when the temperature is 45°C (318.15K)

p = nRT/V

= [ 528 mol C₄H₁₀ · 0.08206 (atm·L/mol·K) · 318.15K ] / 31,200L

= .441 atm = 44.6kPa

ANSWER:

p(T=45°C) = 44.6kPa < p_max = 55.2kPa

Yes, under the conditions listed, the gas cylinder can store gas under 45°C.