Question

If I initially have a gas at a pressure of 12 atm, a volume of 23 liters, and a temperature of 200 K, and then I raise the pressure to 14 atm and increase the temperature to 300 K, what is the new volume of the gas?

Answer 1

30 liters

Step-by-step explanation:

Since a pressure, volume, and temperature is given, this problem involves the combined gas law:

`(P_(1)V_(1) )/(T_(1)) = (P_(2)V_(2) )/(T_(2) )`

1. First, identify the given values. Make sure they are all the same units, and temperature is always in Kelvin.

Your initial values:

P1 = 12 atm

V1 = 23 L

T1 = 200 K

Your final values:

P2 = 14 atm

V2 = ?

T2 = 300 K

2. Now, plug it into the combined gas law formula. If it is easier for you, then you can rearrange the formula for V2.

`(12 * 23)/(200) = (14 * V_(2) )/(300)`

3. Lastly, solve for V2.

`(12 * 23)/(200) = (14 * V_(2) )/(300)`

1.38*300 = 14V2

414 = 14V2

V2 = 29.57142857142857

With sig figs, your final volume would be: 30L

Answer 2

29.57L

Step-by-step explanation:

The following were obtained from the question:

P1 (initial pressure) = 12atm

V1 (initial volume) = 23L

T1 (initial temperature) = 200K

P2 (final pressure) = 14atm

T2 (final temperature) = 300K

V2 (final volume) =?

Using the general gas equation P1V1/T1 = P2V2/T2, the final volume other wise called the new volume of the gas can easily be obtained as follow:

P1V1/T1 = P2V2/T2

12 x 23/200 = 14 x V2/300

Cross multiply to express in linear form as shown below:

200 x 14 x V2 = 12 x 23 x 300

Divide both side by 200 x 14

V2 = (12 x 23 x 300) /(200 x 14)

V2 = 29.57L

Therefore, the new volume of the gas is 29.57L