Question 33: Thermodynamics & KTG

Question

In an adiabatic process, where pressure is decreased by ${3 \over 4}$%, if ${{{C_p}} \over {{C_v}}} = {4 \over 3}$, then the volume increases by

Accepted Answer

In adiabatic process,
pV
$\gamma$
= K
where, K is a constant.
Given that, $\gamma  = {4 \over 3} \Rightarrow p{V^{4/3}} = K$
Taking logarithm on both sides, we get
$\log p + {4 \over 3}\log V = \log K$
$ \Rightarrow {{\Delta p} \over p} + {4 \over 3}{{\Delta V} \over V} = 0$
$ \Rightarrow {{\Delta V} \over V} =  - {3 \over 4}{{\Delta p} \over p}$
Now, % change in volume,
${{\Delta V} \over V} 	imes 100 =  - {3 \over 4}\left( {{{\Delta p} \over p} 	imes 100} ight)$
It is given that, pressure decreases by ${3 \over 4}$%,
i.e. ${{\Delta p} \over p} 	imes 100 =  - {3 \over 4}$
$	herefore$ ${{\Delta V} \over V} 	imes 100 =  - {3 \over 4} 	imes \left( { - {3 \over 4}} ight) = {9 \over {16}}$%
Hence, the volume increases by ${9 \over {16}}$%.

Answer

${3 \over 4}$%

Answer

${9 \over 16}$%

Answer

${16 \over 9}$%

Answer

${4 \over 3}$%

Thermodynamics & KTG

In an adiabatic process, where pressure is decreased by {3 \over 4}%, if {{{C_p}} \over {{C_v}}} = {4 \over 3}, then the volume increases by

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