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1. What is the temperature of 4. 75 moles of a substance at a pressure of 1.2 ATM and a volume of 0.125L? Use 0.08214 R and significant digits (2 in this case).

2. What is the volume of 3.21 moles of a substance at a pressure of 3.4 ATM and a temperature of 29 8L? R equals 0.0821

Explanation:

1)

Given data:

Volume of gas = 0.125 L

Number of moles = 4.75 mol

Temperature of gas =?

Pressure of gas = 1.2 atm

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 put the values in formula:

PV = nRT

T = PV/nR

T = 1.2 atm× 0.125 L /4.75 mol × 0.0821 atm.L/ mol.K

T = 0.15  / 0.39 /K

T = 0.38 K

2)

Given data:

Volume of gas = ?

Number of moles = 3.21 mol

Temperature of gas = 298 K

Pressure of gas = 3.4 atm

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 put the values in formula:

PV = nRT

V = nRT/P

V = 3.21 mol × 0.0821 atm.L/ mol.K× 298 K / 3.4 atm

V = 78.54 L.atm / 3.4 atm

V = 23.1 L

c)red bank

explanation:

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