An ideal gas at 13 degree celsius is heated to double its volume at constant pressure. Find the temperature of the gas
Answers
Answer:
A perfect gas at 27°C is heated at constant pressure till its volume is doubled. What will be its final temperature?
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Using gas equation, we have
(P1*V1) /T1=(P2*V2) /T2 where P=Pressure, V=Volume and T=Absolute temperature of the gas.
Given, P1=P2=P(say) (Since pressure is constant)
V1=V(say), where V is the initial volume of the gas.
Therefore, V2=2V since final volume is double the initial volume.
T1=(27+273)K=300K
T2=?
Substituting the values obtained,in the gas equation, we have
P*V/300=P*2V/T2
Solving, we get T2=final temperature of the gas=600K or (600–273)degree C or 327 degrees C.
Explanation:
Ideal gas equation is pv= mRT ,
In present case , pressure is constant. Thus ideal gas equation can be rewritten as
v/T = mR/p = constant ( since 'mass' and 'characteristic gas constant' are going to be remain same)
Thus,
T2/T1 = v2/v1 ( but in question it was given, v2 = 2×v1)
That is,
T2/(27+273) = 2× v1/v1 ( 273 is added with 27°C to convert in Kelvin)
T2=2×300
T2= 600 k i.e. 327°C( that is your final temperature)
Answer:
299 °C
Explanation:
concept - P1V1/T1=P2V2/T2
when pressure is constant then
V1/T1=V2/T2
given = V2=2V1
T1=13+273= 286K
explanation
V1/286=2V1/T2
T2=2*286
T2= 572K
T2= 572-273= 299 °C
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