Question

What is the volume of the container needed to store 0.8 moles of argon gas at 5.3 atm and 227°C?. . (Given: R = 0.08205 l · atm/mol · K). . 2.81 liters. . 4.39 liters. . 6.19 liters. . 9.67 liters

Answer 1

T = 227 ° C + 273 = 500 K
V = R n T/p = (0.08205 l atm/mol K * 0.8 mol* 500 K) / 5.3 atm = 6.19 lit
Answer: C ) 6.19 liters

Answer 2

Answer:

6.19 L

Step-by-step explanation:

We are given that

Number of moles of argon gas=0.8 moles

Pressure =P=5.3 atm

Temperature =T=227 degree Celsius=$227+273=500k$

We have to find the volume of the container needed to store the argon gas.

We know that

$PV=nRT$

Where R=0.082051 L atm/ mol k

Substitute the values in the given formula

$5.3\times V=0.8\times 500\times 0.82051$

$V=\frac{0.8\times 0.082051\times 500}{5.3}=6.19 L$

Hence, the volume of container needed to store the gas=6.19 L