A 10.3 L cylinder contains 10.67 moles of Cl2 at 21.2 C. What is the pressure (atm)

Explanation:

we know that ideal gas equation is

PV=nRT

P=nRT/V

V=10.3L

n=10.67

T=21.2+273=294

R=0.0821

P=10.67*0.0821*294/10.3

P=25.00atm

P = 25.0 atm

Explanation:

Given data:

Volume of gas = 10.3 L

Number of moles of Cl₂ = 10.67 mol

Temperature = 21.2 °C

Pressure of gas = ?

Solution:

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

Now we will convert the temperature.

21.2+273 = 294.2 K

10.3 L × P = 10.67 mol ×  0.0821 atm.L/ mol.K  × 294.2 K

P = 257.72 atm.L /  10.3 L

P = 25.0 atm

just did the assignment and the answer was #1 +2 n