Questions
The volume of a gas is reduced adiabatically to of its volume at = 1.4, the new temperature will be
a
150 × (4)0.4K
b
300 × (4)0.4K
c
250 × (4)0.4 K
d
none of these
detailed solution
Correct option is B
For adiabatic change the relation between temperature and volume isTVγ-1 = constantwhere γ is ratio of specific heats of the gas,Given, T1 = 27 + 273 = 300 K, V1 = V, V2 = V4T1Vγ-11 = T2Vγ-12⇒ T2 = (V1V2)γ-1×T1 = (VV4)1.4-1×300 = (4)0.4×300 = 300 × (4)0.4KTalk to our academic expert!
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During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its temperature. The ratio of for the gas is
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