Answer:
the final temperature is T2= -126.5°C
Explanation:
Since the gas behaves as an ideal gas
Initial state) P1*V1=n*R*T1
Final state) P2*V2=n*R*T2
where P= gas pressure , V=volume occupied by the gas , n= number of moles , R= ideal gas constant , T= absolute temperature
dividing both equations
(P2/P1)*(V2/V1)=T2/T1
since P2=2*P1 and V2=1/4*V1 , T1=20°C=293K
2*1/4=T2/T1
T2=T1/2 = 293K/2 = 146.5K
T2= 146.5K= -126.5°C
The final temperature of the gas will be [tex]T_{f} =-126.5 C[/tex]
It is given that
Initial temperature = 20 C = 273+20=239k
Now the pressure is doubled so new pressure
[tex]P_{n} = 2P[/tex]
Volume is reduced to one fourth then
[tex]V_{n} =\dfrac{V}{4}[/tex]
Now we know that from the ideal gas equation
The initial condition of the gas
[tex]PV=nRT[/tex]
[tex]PV=nR(293)[/tex].......................(1)
For the final condition of the gas
[tex]P_{n} V_{n} =nRT_{n}[/tex]
putting values
[tex]2P\times \dfrac{V}{4} } =nRT_{n}[/tex]
[tex]PV=2nRT_{n}[/tex].............................(2)
from equations (1) and (2) we get
[tex]2nRT_{n} =nR(293)[/tex]
[tex]T_{n} = 146.5-273=-126.5 C[/tex]
Thus the final temperature of the gas will be [tex]T_{f} =-126.5 C[/tex]
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