Question 29: 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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