Question #22695

How moles of oxygen are there in 50L tank at 21 c when the oxygen pressure is 15.7atm?

Expert's answer

pV = nRT

, where P is the pressure of the gas, V is the volume of the gas, n is the amount of substance of gas (also known as number of moles), T is the temperature of the gas and R is the ideal, or universal, gas constant , equal to the product of Boltzmann's constant and Avogadro's constant. Hence,

n = pV/RT

p = 15.7 atm

1 atm = 1.013 * 10^5 Pa

p = 15.7*1.013 *10^5 = 1.59 * 10^6 Pa

V = 50 * 10^-3 m^{3}

T = 294 K

R = 8.314 J/mol*K

n = (1.59 * 10^6 Pa * 50 * 10^-3 m3)/(8.314 J/mol*K * 294K) = 35.52 mol

Answer:

n(H_{2}) = 32.52 mol

, where P is the pressure of the gas, V is the volume of the gas, n is the amount of substance of gas (also known as number of moles), T is the temperature of the gas and R is the ideal, or universal, gas constant , equal to the product of Boltzmann's constant and Avogadro's constant. Hence,

n = pV/RT

p = 15.7 atm

1 atm = 1.013 * 10^5 Pa

p = 15.7*1.013 *10^5 = 1.59 * 10^6 Pa

V = 50 * 10^-3 m

T = 294 K

R = 8.314 J/mol*K

n = (1.59 * 10^6 Pa * 50 * 10^-3 m3)/(8.314 J/mol*K * 294K) = 35.52 mol

Answer:

n(H

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