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Question: A gas at \[{10^ \circ }C\] temperature and \[1.013{\text{ }} \times {10^3}pa\] pressure is compresse...

A gas at 10∘C{10^ \circ }C temperature and 1.013 ×103pa1.013{\text{ }} \times {10^3}pa pressure is compressed adiabatically to half of its volume. If the ratio of specific heat of the gas is 1:4 , what is its final temperature?

Explanation

Solution

First we will substitute the values of the given question into the formula of adiabatic compression to give an answer . we must remember to convert the final answer into ∘C^ \circ Cfrom kelvin. We will learn a little about specific heats and adiabatic compression.
Formula used:
PVγ=kP{V^\gamma } = k
Where P=pressure,
V=volume and
k= constant
γ\gamma =ratio of specific heat.

Complete answer:
Let us first know what adiabatic compression is: An adiabatic process is a sort of thermodynamic process that happens without the passage of heat or mass between the thermodynamic system and its environment in thermodynamics. An adiabatic process, unlike an isothermal process, only sends energy to the environment as work.
Specific heats: The specific heat capacity of a material is defined as the heat capacity of a sample of the substance divided by the mass of the sample in thermodynamics. Specific heat capacity is also known as massic heat capacity.
Looking at all the given information,
Let's substitute in formula:
PVγ=kP{V^\gamma } = k
Or, TVγ−1=kT{V^{\gamma - 1}} = k
Or, T1V1γ−1=T2V2γ−1{T_1}{V_1}^{\gamma - 1} = {T_2}V_2^{\gamma - 1}
Or, T2=T1V1γ−1V2γ−1{T_2} = {T_1}\dfrac{{{V_1}^{\gamma - 1}}}{{V_2^{\gamma - 1}}}
Or, T2=(10+273)(V1V2)γ−1{T_2} = \left( {10 + 273} \right){\left( {\dfrac{{{V_1}}}{{{V_2}}}} \right)^{\gamma - 1}}
Or, T2=283(2)1.4−1{T_2} = 283{\left( 2 \right)^{1.4 - 1}}
(because it is given that volume is reduced to half hence V1V2=2\dfrac{{{V_1}}}{{{V_2}}} = 2 and value of γ=1.4\gamma = 1.4)
Or, T2=283×20.4K{T_2} = 283 \times {2^{0.4}}K
OR, T2=373.56K{T_2} = 373.56K
Or, T2=(373.56−273)∘C{T_2} = {\left( {373.56 - 273} \right)^ \circ }C
Or, T2=100.56∘C{T_2} = {100.56^ \circ }C
Hence the correct answer for the question is T2=100.56∘C{T_2} = {100.56^ \circ }Cor its final temperature is T2=100.56∘C{T_2} = {100.56^ \circ }C.

Note:
Usually mistakes are made while writing the formula for adiabatic compression. Remember that it is TVγ−1=kT{V^{\gamma - 1}} = k and not TVγ=kT{V^\gamma } = k.