Answer to Question #343992 in Chemistry for Rey

Question

Solve the following problems using the correct format. Given first, what is asked for, and the formula together with the solution. 

1. A gas occupies a volume of 400 cm³ at a pressure of 2.00 atm. What would be the new pressure if the volume is increased to 500 cm³ ?
2. The volume of a given gas was 50 cm3 at 260C. After heat was applied to the gas, its volume increased to 80 cm3 . If the pressure was held constant, what is the new temperature of the gas? 

Answer 1

1. A gas occupies a volume of 400 cm³ at a pressure of 2.00 atm. What would be the new pressure if the volume is increased to 500 cm³?

Given:

P₁ = 2.00 atm

V₁ = 400 cm³

T = constant

P₂ = unknown

V₂ = 500 cm³

Formula: P₁V₁ = P₂V₂

Solution:

Since the temperature and amount of gas remain unchanged, Boyle's law can be used.

Boyle's gas law can be expressed as: P₁V₁ = P₂V₂

To find the new pressure, solve the equation for P₂:

P₂ = P₁V₁ / V₂

P₂ = (2.00 atm × 400 cm³) / (500 cm³) = 1.60 atm

P₂ = 1.60 atm

Answer: The new pressure will be 1.60 atm

2. The volume of a given gas was 50 cm³ at 26°C. After heat was applied to the gas, its volume increased to 80 cm³. If the pressure was held constant, what is the new temperature of the gas? 

Given:

V₁ = 50 cm³

T₁ = 26°C + 273.15 = 299.15 K

P = constant

V₂ = 80 cm³

T₂ = unknown

Formula: V₁/T₁ = V₂/T₂

Solution:

Since the pressure and amount of gas remain unchanged, Charles's law can be used.

Charles's law can be expressed as: V₁ / T₁ = V₂ / T₂

V₁T₂ = V₂T₁

To find the new temperature of the gas, solve the equation for T₂:

T₂ = V₂T₁ / V₁

T₂ = (80 cm³ × 299.15 K) / (50 cm³) = 478.64 K

T₂ = 478.64 K (or 205.5°C)

Answer: The new temperature of the gas will be 478.64 K (or 205.5°C)