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The volume of a gas is halved during an adiabatic compression that increases the pressure by a factor of 2.5. By what factor does the temperature increase?

Sagot :

Answer: 1.24

Explanation:

Given

Volume becomes half and pressure increases by a factor of 2.5

In adiabatic process [tex]PV^{\gamma}=\text{constant}\ \text{or}\ TV^{\gamma}=\text{constant}[/tex]

Finding out [tex]\gamma[/tex] first

[tex]\Rightarrow PV^{\gamma}=2.5P(0.5V)^{\gamma}\\\\\Rightarrow \left(\dfrac{V}{0.5V}\right)^{\gamma}=2.5\\\\\Rightarrow 2^{\gamma}=2.5\\\\\text{Taking log both side}\\\\\Rightarrow \gamma=\dfrac{\ln (2.5)}{\ln (2)}\\\\\Rightarrow \gamma=1.32[/tex]

Applying same principle for Temperature

[tex]\Rightarrow TV^{1.32-1}=T'(0.5V)^{1.32-1}\\\\\Rightarrow T'=(2)^{0.32}T\\\\\Rightarrow T'=1.24T[/tex]

Thus, the temperature increases by a factor of [tex]1.24[/tex]

The factor by which the temperature increases is; 1.24

How to find increase in temperature in an adiabatic process?

In thermodynamics, an adiabatic process is one that happens when there is zero heat transfer between the system and its environment. Thereafter, the internal energy change becomes the total work done by the system. The formula associated with this process for an ideal gas is;

[tex]PV^{\gamma }[/tex] = Constant

We are told that there is an increase in pressure by a factor of 2.5 and that the volume is halved. Thus, we will have;

[tex]PV^{\gamma} = 2.5P(0.5V)^{\gamma}[/tex]

This will be simplified to;

[tex](\frac{1}{0.5})^{\gamma} = 2.5[/tex]

⇒ [tex]\gamma = \frac{In2.5}{In 2}[/tex]

γ = 1.32

Now, If the volume expands in an adiabatic process, then the volume temperature relation is expressed as:

[tex]T_{1}V_{1}^{\gamma - 1 } = T_{2}V_{2}^{\gamma - 1 }[/tex]

[tex]T_{1}V_{1}^{1.32 - 1 } = T_{2}(0.5V_{1})^{1.32 - 1 }[/tex]

Simplifying this gives;

T₁ = 1.24T₂

Read more about Adiabatic Process at; https://brainly.com/question/3962272