Question

an open container of volume 3L contains air at 1atm. The container is heated from 27°C to t°C. The amount of gas expelled from the container measured 1.45L at 17°C at 1atm. Find t. ​

Answer 1

Answer: According to the ideal gas equation

PV=nRT

nR is constant

so PV/T before the expulsion= PV/T after the expulsion

volume of gas in the container = 3-1.45

                                                   =2.55

1x3/300 = 1x2.55/t

t=255 K

In celsius

255-273 = -18C.

Explanation:

Answer 2

Using the combined gas equation
P1V1/T1 =P2V2/T2
Therefore substitute the values,
V1 =3l,p1=1atm,t1=t-27 degrees Celsius which is equal to t+273-300 degrees in Kelvin scale.
V2=1.45l,p2=1atm, t2=17 degrees Celsius which is equal to 290 kelvin.
Therefore by the combined gas equation
3*1/t-27=1.45*1/290
3/t-27=1/200
t-27/3=200
t-27=600
T=627K which is equal to 354 degrees Celsius
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