Question

A 400 ml sample of gas is heated from -20 c to 60

c. the pressure changes from 50 to 225 kpa. what is the final vollume

Answer 1

using p1v1/t1=p2v2/t2
p1=50
p2=225
v1=400ml
v2=?
t1=-20=253k
t2=60=333k
50x400/253=225xv2/333
7.9=0.7xv2
v2=7.9/0.7
v2=11.3ml

Answer 2

Answer : The new volume of gas will be 117 ml.

Step-by-step explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

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

where,

`P_1` = initial pressure of gas = 50 kPa

`P_2` = final pressure of gas = 225 kPa

`V_1` = initial volume of gas = 400 ml

`V_2` = final volume of gas = ?

`T_1` = initial temperature of gas =
`-20^0C=(-20+273)K=253K`

`T_2` = final temperature of gas =
`60^0C=(60+273)K=333K`

Now put all the given values in the above equation, we get the final pressure of gas.

`(50* 400)/(253K)=(225* V_2)/(333K)`

`V_2=117ml`

Therefore, the new volume of gas will be 117 ml.