Question

a mono atomic gas initially at $27^oc$is compressed adiabatically to one eighth of its original volume .the temperature after compression will be?

Answer 1

Answer:

10∘C

887∘C

927∘C

144∘C

Answer :

C

Solution :

T1Vγ−11=T2Vγ−12

Answer 2

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 :

$\longmapsto\bold{T_1 V_1^{\ \gamma - 1} = T_2 V_2^{\ \gamma - 1} }$

Here, as the gas is monoatomic , so γ = 5/3

Now plugging in all values :

$\longmapsto\sf 300 \times V_1^{\ \big(\dfrac{5}{3} - 1\big) } = T_2 \times \bigg(\dfrac{V_1}{8} \bigg)^{ \big(\dfrac{5}{3} - 1 \big)} \\\\\\ \longmapsto\sf 300 \times V_1^{\big(\dfrac{2}{3}\big)} = T_2 \times \bigg(\dfrac{V_1}{8}\bigg)^{\big(\dfrac{2}{3}\big)} \\\\\\ \longmapsto\sf \dfrac{300}{T_2} = \bigg(\dfrac{ \dfrac{V_1}{8}}{V_1}} \bigg)^{\dfrac{2}{3}} \\\\\\ \longmapsto\sf \dfrac{300}{T_2} = \bigg(\dfrac{1}{8} \bigg)^{\dfrac{2}{3}}$

$\longmapsto\sf \dfrac{300}{T_2} = \bigg(\dfrac{1}{2}\bigg)^{3 \times \dfrac{2}{3}} \\\\\\ \longmapsto\sf \dfrac{300}{T_2} = \bigg( \dfrac{1}{2} \bigg)^{2} \\\\\\ \longmapsto\sf \dfrac{300}{T_2} = \dfrac{1}{4} \\\\\\ \longmapsto\sf T_2 = 300 \times 4 \\\\\\ \longmapsto\underline{\boxed{\sf T_2 = 1200 \ K }}} \\\\\\ \longmapsto\sf T_2 = (1200 - 273)^{\circ} \ C \\\\\\ \longmapsto\underline{\underline{\boxed{\sf{T_2 = 927^{\circ} \ C }}}}$

∴ Temperature after compression = 1200K or 927°C