Question

Given 4.8 moles of a gas in a 1.8 L 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 and 1 atm = 760 torr.)
6.79 torr
66.55 torr
6.79 x 100 atm
6.66 x 101 atm

Answer 1

PV = nRT equation to use

P(1.8) = 4.8(0.0821)(304)
P(1.8) = 119.8
P = 66.55 atm or 6.66x10^1 atm

Answer 2

Answer : The pressure of the gas is, $6.66\times 10^{1}atm$

Solution :

Using ideal gas equation,

$PV=nRT$

where,

n = number of moles of gas  = 4.8 moles

P = pressure of the gas = ?

T = temperature of the gas = $31^oC=273+31=304K$

R = gas constant = 0.0821 Latm/moleK

V = volume of gas = 1.8 L

Now put all the given values in the above equation, we get the pressure of the gas.

$P\times (1.8L)=4.8moles\times 0.0821L.atm/mole.K\times 304K$

$P=66.6atm=6.66\times 10^{1}atm$

Therefore, the pressure of the gas is, $6.66\times 10^{1}atm$