A diatomic ideal gas is compressed adiabatically to $\frac{1}{32}$ of its initial volume. If the initial temperature of the gas is $T_{i}$ (in Kelvin) and the final temperature is $a T_{i}$ , the value of a is
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Question

A diatomic ideal gas is compressed adiabatically to $\frac{1}{32}$ of its initial volume. If the initial temperature of the gas is $T_{i}$ (in Kelvin) and the final temperature is $a T_{i}$ , the value of a is
4
2
3
5

Toppr Admin · Accepted Answer

For adiabatic process- we can write TV-x3B3-x2212-1- constant-x3B3-75 for diatomic gasesTiV75-x2212-1- k-constant- -initial conditionaTiV75-x2212-1- k-constant- -final conditionTV75-x2212-1-aT-V32-75-x2212-1-x2234-a-4

A diatomic ideal gas is compressed adiabatically to 132 of its initial volume. If the initial temperature of the gas is Ti (in Kelvin) and the final temperature is aTi, the value of a is4235

For adiabatic process, we can write TVγ−1= constantγ=75 for diatomic gasesTiV75−1= k(constant) -------initial conditionaTiV75−1= k(constant) ------final conditionTV75−1=aT(V32)75−1∴a=4

A diatomic ideal gas is compressed adiabatically to $\frac{1}{32}$ of its initial volume. If the initial temperature of the gas is $T_{i}$ (in Kelvin) and the final temperature is $a T_{i}$ , the value of a is
4
2
3
5