Question

When the temperature of a closed vessel containing a gas is increased by 273 °C, the pressure was found to increase by 20%. What is the initial temperature of the closed vessel?


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Answer 1

Hey There!

The answer is 1365k

Step-by-step explanation:

Refer to the attached picture.

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Answer 2

Solution given;

1st temperature [t1]=T

initial temperature [t2]=T+273°C

1st pressure [P]=P

and

initial pressure[P2]=P+20%of P=120/100V=1.2P

now

we have

According to Gay lussac's law

P
`\alpha` T

`(P)/(T)`=K where k is constant

at 2 different states

`(P1)/(T1)`=
`(P2)/(T2)`

• 
  `(P)/(T)`=
  `(1.2P)/(T+273)`

`(T+273)/(T)`=
`(1.2P)/(P)`

doing crisscrossed multiplication

T+273=1.2T

273=1.2T-T

0.2T=273

T=
`(273)/(0.2)`=1365°C or 1365K

Initial temperature of the closed vessel is 1365°C or 1365K.