Question

A container holds 35.8 moles of gas under 10.0 atm of pressure at 70.0 C. What is the volume of the container? 20.6 L 101 L 2080 L 10200 L

Answer 1

PV = n RT
P: pressure =10atm
V volume
n number of mole = 35.8 moles
R universal gas constant = 0.082
T: The temperature= 70°C= 343.15 Kelvin

 
V= (n RT) / P = 35.8 x 0.082 x 343.15   / 10 = 100.7 ≈ 101 L

 V = 101L

Answer 2

Answer : The volume of the gas is, 101 liters

Solution :

Using ideal gas equation :

$PV=nRT\\\\V=\frac{nRT}{P}$

where,

n = number of moles of gas  = 35.8 moles

P = pressure of the gas = 10.0 atm

T = temperature of the gas = $70^oC=273+70=343K$

R = gas constant = 0.0821 L.atm/mole.K

V = volume of gas = ?

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

$V=\frac{nRT}{P}$

$V=\frac{35.8mole\times (0.0821L.atm/mole.K)\times 343K}{10atm}$

$V=100.81L\approx 101L$

Therefore, the volume of the gas is, 101 liters