a mono atomic gas initially at [tex]27^oc[/tex]is compressed adiabatically to one eighth of its original volume .the temperature after compression will be?
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a mono atomic gas initially at [tex]27^oc[/tex]is compressed adiabatically to one eighth of its original volume .the temperature after compression will be?
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Answer:
10∘C
887∘C
927∘C
144∘C
Answer :
C
Solution :
T1Vγ−11=T2Vγ−12
Verified answer
Question :
A mono atomic gas initially at 27°C is compressed adiabatically to one eighth of its original volume .the temperature after compression will be?
Solution :
Given , initial temperature , T₁ = 27° C = (273 + 27) = 300K
Initial volume, V = V₁
Final volume , V₂ = V₁/8
Final temperature , T₂ = ?
Formula used :
[tex]\longmapsto\bold{T_1 V_1^{\ \gamma - 1} = T_2 V_2^{\ \gamma - 1} }[/tex]
Here, as the gas is monoatomic , so γ = 5/3
Now plugging in all values :
[tex]\longmapsto\sf 300 \times V_1^{\ \big(\dfrac{5}{3} - 1\big) } = T_2 \times \bigg(\dfrac{V_1}{8} \bigg)^{ \big(\dfrac{5}{3} - 1 \big)} \\\\\\ \longmapsto\sf 300 \times V_1^{\big(\dfrac{2}{3}\big)} = T_2 \times \bigg(\dfrac{V_1}{8}\bigg)^{\big(\dfrac{2}{3}\big)} \\\\\\ \longmapsto\sf \dfrac{300}{T_2} = \bigg(\dfrac{ \dfrac{V_1}{8}}{V_1}} \bigg)^{\dfrac{2}{3}} \\\\\\ \longmapsto\sf \dfrac{300}{T_2} = \bigg(\dfrac{1}{8} \bigg)^{\dfrac{2}{3}}[/tex]
[tex]\longmapsto\sf \dfrac{300}{T_2} = \bigg(\dfrac{1}{2}\bigg)^{3 \times \dfrac{2}{3}} \\\\\\ \longmapsto\sf \dfrac{300}{T_2} = \bigg( \dfrac{1}{2} \bigg)^{2} \\\\\\ \longmapsto\sf \dfrac{300}{T_2} = \dfrac{1}{4} \\\\\\ \longmapsto\sf T_2 = 300 \times 4 \\\\\\ \longmapsto\underline{\boxed{\sf T_2 = 1200 \ K }}} \\\\\\ \longmapsto\sf T_2 = (1200 - 273)^{\circ} \ C \\\\\\ \longmapsto\underline{\underline{\boxed{\sf{T_2 = 927^{\circ} \ C }}}}[/tex]
∴ Temperature after compression = 1200K or 927°C