calculate the pressure in a 212 liter tank containing 23.3 kg of argon gas at 25°c
Answers
Answer:
First of all, let's find the number of moles of the gas.
The molar mass of argon is M_m=40 g/mol=0.40 kg/molM
m
=40g/mol=0.40kg/mol . Since we have m=23.3 kgm=23.3kg of gas, the number of moles is
n= \frac{m}{M_m}= \frac{23.3 kg}{0.40 kg/mol}=58.3 moln=
M
m
m
=
0.40kg/mol
23.3kg
=58.3mol
Now we can use the ideal gas law to calculate the pressure of the gas:
pV=nRTpV=nRT
where
p is the pressure
V=212 L=0.212 m^3V=212L=0.212m
3
is the volume
n=58.3 moln=58.3mol is the number of moles
R=8.31 J/mol KR=8.31J/molK is the gas constant
T=25^{\circ}+273=298 KT=25
∘
+273=298K is the absolute temperature
Rearranging the equation, we find
p= \frac{nRT}{V}= \frac{(58.3 mol)(8.31 J/mol K)(298 K)}{0.212 m^3}=6.81 \cdot 10^5 Pap=
V
nRT
=
0.212m
3
(58.3mol)(8.31J/molK)(298K)
=6.81⋅10
5
Pa