Question

Calculate the final temperature of a monoatomic ideal gas that os compressed reversibly and adiabatically from 16l to 2l at 300k

Answer 1

The final temperature of a monoatomic ideal gas that is compressed reversibly and adiabatically from 16 L to 2 L at 300 K temperature is 1200 K.

Given-

• Initial volume = 16 L
• Final volume = 2 L
• Temperature = 300 K

In case of reversible adiabatic

TV^γ-1 = constant where T is the temperature, V is the volume and γ is the ratio of specific heats at constant pressure and at constant volume.

So,

T₁V₁^ γ-1 = T₂V₂^γ -1

By substituting the values we get

T₂/T₁ = (V₁/V₂)^γ -1

We know that for monoatomic gases γ  is 5/3. So,

T₂ = 300 × (16/2)^5/3-1

T₂ = 1200 K

Answer 2

Given :

• A monoatomic Ideal gas expands adiabatically
• $\sf\:V_{1}=16\:l$
• $\sf\:V_{2}=2\:l$
• $\sf\:T_{1}=300\:K$

To find :

Final temperature ,$\sf\:T_{2}$

Theory:

•The relation between temperature T and volume V from the adiabatic gas equation is

$\bf\:TV{}^{\gamma-1}=Costant$

•Degree of freedom for monoatomic gases= 3

$\sf\:C_{v}=\dfrac{3}{2}R$

and $\sf\:C_{p}=\dfrac{5}{2}R$

$\implies\:\sf\:\gamma\:=\dfrac{C_{p}}{C_{v}}=\dfrac{5}{3}$

Solution :

We know that

$\bf\:TV{}^{\gamma-1}=Costant$

$\implies\:\sf\:T_{1}V{}^{\gamma-1}_{1}=T_{2}V{}^{\gamma-1}_{2}$

$\implies\:\sf\:T_{2} =T_{1} \times ( \frac{V_{1}}{V_{2}} ) {}^{ \gamma - 1}$

Now put the given values.

For Monoatomic gases

$\sf\:\gamma=\dfrac{5}{3}$

$\implies\:\sf\:T_{2} = 300 \times ( \frac{16}{2} ) {}^{ \frac{5}{3} - 1}$

$\implies\:\sf\:T_{2} = 300(8) {}^{ \frac{2}{3} }$

$\implies\:\sf\:T_{2} = 300 \times 2 {}^{3 \times } {}^{ \frac{2}{3} }$

$\implies\:\sf\:T_{2} = 300 \times 4 = 1200k$