Question

The pressure P of a gas varies directly with the temperature T and inversely with the volume V. A certain gas has a volume of 6 liters (L), a temperature of 400 Kelvins (K), and a pressure of 2.4 atmospheres (atm). If the gas is expanded to a volume of 15 L and is heated to 650 K, what will the new pressure be?

Answer 1

1.56 atm

Step-by-step explanation:

Step 1 Find k.

V = kT/P

6 = K(400)/2.4 Substitute.

k = 0.036 Solve for k.

Step 2 Use the variation function.

V = 00.36T/P Use 0.036 for k.

15 = 00.36T(650)/P Substitute.

15 = 23.4/P Simplify.

P = 1.56 Solve for P.

The new pressure will be 1.56 atm.

Answer 2

The value of new pressure will be 1.56 atm.

Step-by-step explanation:

Using combined gas law :

The combined gas equation is,

`(P_1V_1)/(T_1)=(P_2V_2)/(T_2)`

where,

`P_1` = initial pressure of gas = 2.4 atm

`P_2` = final pressure of gas = ?

`V_1` = initial volume of gas =
`6 L`

`V_2` = final volume of gas = 15 L

`T_1` = initial temperature of gas =
`400 K`

`T_2` = final temperature of gas =
`650 K`

Now put all the given values in the above equation, we get:

`(2.4 atm * 6 L)/(400 K)=(P_2* 15 L)/(650 K)`

`P_2=(2.4 atm * 6 L* 650 K)/(400 K* 15 L)`

`P_2=1.56 atm`

The value of new pressure will be 1.56 atm.