A thermally insulated vessel contains an ideal gas of molecular mass M and ratio of specific heats 1.4. Vessel is moving with speed v and is suddenly brought to rest. Assuming no heat is lost to the surrounding and vessel temperature of the gas increases by :
(R = universal gas constant)
Solution
<p>Let there be n moles of gas</p>
<p>E<sub>loss</sub> = E<sub>gain</sub></p>
<p>${1 \over 2}(nM){v^2} = n{C_v}\Delta T$</p>
<p>${1 \over 2}M{v^2} = {C_v}\Delta T$</p>
<p>here, $\gamma = 1.4 = {7 \over 5}$ i.e. diatomic gas</p>
<p>$\therefore$ ${C_v} = {{5R} \over 2}$</p>
<p>Now, ${1 \over 2}M{v^2} = {{5R} \over 2}\Delta T$</p>
<p>$\Delta T = {{M{v^2}} \over {5R}}$</p>
About this question
Subject: Physics · Chapter: Thermodynamics · Topic: Zeroth and First Law
This question is part of PrepWiser's free JEE Main question bank. 271 more solved questions on Thermodynamics are available — start with the harder ones if your accuracy is >70%.