Question

A gas takes up a volume of 17.0 L, has a pressure of 2.30 atm, and a temperature of 299 K. If the temperature is raised to 350.0 K and the pressure lowered to 1.50 atm, what is the new volume of the gas?

Answer 1

Answer:

3.03 L

Explanation:

Using the combined gas law which states that the volume (V) of an amount of gas is directly proportional to the temperature (T) of the gas and inversely proportional to the pressure (P). i.e

V ∝ T / P

=> PV / T = K

=> P₁ V₁ / T₁ = P₂ V₂ / T₂        --------------------------(i)

Where;

P₁ and P₂ are initial and final values of the pressure

V₁ and V₂ are initial and final values of the volume

T₁ and T₂ are initial and final values of the temperature

From the question, the following are given;

P₁ = 2.30 atm

P₂ = 1.50 atm

V₁ = 17.0 L

T₁ = 299 K

T₂ = 350.0 K

To calculate the new volume (V₂) substitute the values above into equation (i) as follows;

=> 2.30 x 17.0 / 299 = 1.50 x V₂ / 350.0

=> 0.013 = 0.00429 x V₂

Solve for V₂;

=> V₂ = 0.013 / 0.00429

=> V₂ = 3.30 L

Therefore, the new volume of the gas is 3.03L

Answer 2

Answer:

new volume = 30.513L

Explanation:

V1 =17L , P1= 2.3atm ,  T1=299K ,  T2=350K , P2 =1.5atm,   V2=?

To find the new volume of the gas we will use the combined gas law.

The combined gas law combines Boyle's law, Charles law and Gay-Lussac's law. Its states that the ratio of the product of pressure and volume to temperature is equal to a constant. its expressed mathematically below.

$\frac{V1P1}{T1} = \frac{V2P2}{T2}$

$V2 =\frac{V1P1T2}{P2T1}$

$V2 =\frac{17*2.3*350}{1.5*299}$

$v2 = \frac{13685}{448.5}$

V2 =30.513L