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[tex] \alpha [/tex] T
[tex] \frac{P}{T} [/tex]=K where k is constant
at 2 different states
[tex] \frac{P1}{T1} [/tex]=[tex] \frac{P2}{T2} [/tex]
- [tex] \frac{P}{T} [/tex]=[tex] \frac{1.2P}{T+273} [/tex]
[tex] \frac{T+273}{T} [/tex]=[tex] \frac{1.2P}{P} [/tex]
doing crisscrossed multiplication
T+273=1.2T
273=1.2T-T
0.2T=273
T=[tex] \frac{273}{0.2} [/tex]=1365°C or 1365K
Initial temperature of the closed vessel is 1365°C or 1365K.
