Question

Given 2.91 moles of a gas in a 500 milliliter-container, if the temperature is found to be 31 degrees Celsius, what is the pressure of the gas? (The ideal gas constant is 0.0821 L · atm/mol · K.)
1.45 x 102 atm
1.48 x 101 atm
1.45 x 10-1 atm
1.45 x 101 atm

Answer 1

Answer: The pressure of the gas comes out to be $1.45\times 10^2atm$

Explanation:

To calculate the pressure of the gas, we use the equation given by ideal gas equation, which is:

$PV=nRT$

where,

P = Pressure of the gas = ? atm

V = Volume of the gas = 500 mL = 0.5 L    (Conversion factor: 1 L = 1000 mL)

n = Number of moles of gas = 2.91 moles

R = Gas constant = $0.0821\ttext{ L atm }mol^{-1}K^{-1}$

T = Temperature of the gas = $31^oC=[273+31]K=304K$

Putting values in above equation, we get:

$P\times 0.5L=2.91mol\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 304K\\\\P=145.2atm=1.45\times 10^2atm$

Hence, the pressure of the gas comes out to be $1.45\times 10^2atm$

Answer 2

Use the PV = nRT equation T is in Kelvins = 31 + 273 = 304 K

P(0.5) = (2.91)(0.0821)(304)
P(0.5) = 72.6289
P = 145.25 atm or 1.45x10^2 atm