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?

No spamming or Irrelevant Answer.
Ty!​

Answer 1

Answer:

Hey There!

The answer is 1365k

Explanation:

Refer to the attached picture.

Hope it Helps:)

~ƒöᏝϗʆѻʁᶥąռ¹₃

Answer 2

Answer:

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

$\frac{P}{T}$=K where k is constant

at 2 different states

$\frac{P1}{T1}$=$\frac{P2}{T2}$

• $\frac{P}{T}$=$\frac{1.2P}{T+273}$

$\frac{T+273}{T}$=$\frac{1.2P}{P}$

doing crisscrossed multiplication

T+273=1.2T

273=1.2T-T

0.2T=273

T=$\frac{273}{0.2}$=1365°C or 1365K

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