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On increasing the temperature of a monoatomic ideal gas adiabatically by 1°C, the volume decreases by 1.5%. Then the initial temperature of gas is :

On increasing the temperature of a monoatomic ideal gas adiabatically by 1°C, the volume decreases by 1.5%. Then the initial temperature of gas is :

Grade:12

1 Answers

Sunil Kumar FP
askIITians Faculty 183 Points
9 years ago
for monoatomic ideal gas during adiabatic process,
we have TV^(g-1)=constant
or T1/T2= (V2/V1)^g-1
g is the ratio of specific heat capacity for monoatomic gas.g=1.6
here the initial temperature=T K
final temperature=T+1 K
initial volume =100 Lt suppose
final volume =98.5L
Thus applying the above equation
T/(T+1)=(98.5/100)^1.6-1
T/(T+1)=.985^.6
T/T+1 =.9909
T=.9909T+.9901
.009027T= .9901
T=108.89K

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