Question

A gas at 300 k and 4.0 atm is moved to a new location with a temperature of 250 k. the volume changes from 5.5 l to 2.0 l. what is the pressure of the gas at the new location? use the formula: p1v1 t1 = p2v2 t2

Answer 1

Answer:

9.17 atm

Explanation:

To find the new pressure of the gas, you need to use the following manipulated formula:

P₁V₁ / T₁ = P₂V₂ / T₂

In this formula,

P₁ = initial pressure (atm)              P₂ = new pressure (atm)

V₁ = initial volume (L)                    V₂ = new volume (L)

T₁ = initial temperature (K)            T₂ = new temperature (K)

Because you have been given values for all of the variables except for the new pressure, you can substitute them into the equation and simplify.

P₁ = 4.0 atm                                  P₂ = ? atm

V₁ = 5.5 L                                      V₂ = 2.0 L

T₁ = 300 K                                     T₂ = 250 K

P₁V₁ / T₁ = P₂V₂ / T₂                                                     <----- Given formula

(4.0 atm)(5.5 L) / (300 K) = P₂(2.0 L) / (250 K)           <----- Insert variables

0.073333 = P₂(2.0 L) / (250 K)                                   <----- Simplify left side

18.33333 = P₂(2.0 L)                                     <----- Multiply both sides by 250

9.17 = P₂                                                        <----- Divide both sides by 2.0

Answer 2

Answer:

=> 13.2 atm

Explanation:

From the question, we have question we have been provided with pressure, volume and temperature. Based on these variables, we are supposed to use the combined gas law formula (P1V1T1 = P2V2T2)

                   At location 1;

P1 = 4.0 atm

V1 =  5.5 L

T1 = 300 K.

                       At location 2;

P2 = ?

V2 = 2.0 L

T2 = 250 K.

We are required to determine the pressure (P2).

To begin, we make pressure (P2) the subject of in the formula:

      P2 = (P1V1T1)/(V2T2)  ................ eq. (i)

Then substitute the known values into the eq. (i).

  P2 = (4.0 x 5.5 x 300)/(2.0 x 250) atm

       = 13.2 atm.

Therefore, the pressure (P2) at location 2 is 13.2 atm.