JEE MAIN - Physics (2011 - No. 18)
A thermally insulated vessel contains an ideal gas of molecular mass $$M$$ and ratio of specific heats $$\gamma .$$ It is moving with speed $$v$$ and it's suddenly brought to rest. Assuming no heat is lost to the surroundings, Its temperature increases by:
$${{\left( {\gamma - 1} \right)} \over {2\gamma R}}M{v^2}K$$
$${{\gamma {M^2}v} \over {2R}}K$$
$${{\left( {\gamma - 1} \right)} \over {2R}}M{v^2}K$$
$${{\left( {\gamma - 1} \right)} \over {2\left( {\gamma + 1} \right)R}}M{v^2}K$$
स्पष्टीकरण
Here, work done is zero.
So, loss in kinetic energy $$=$$ change in internal energy of gas
$${1 \over 2}m{v^2} = n{C_v}\Delta T = n{R \over {\gamma - 1}}\Delta T$$
$${1 \over 2}m{v^2} = {m \over M}{R \over {\gamma - 1}}\Delta T$$
$$\therefore$$ $$\Delta T = {{M{v^2}\left( {\gamma - 1} \right)} \over {2R}}K$$
So, loss in kinetic energy $$=$$ change in internal energy of gas
$${1 \over 2}m{v^2} = n{C_v}\Delta T = n{R \over {\gamma - 1}}\Delta T$$
$${1 \over 2}m{v^2} = {m \over M}{R \over {\gamma - 1}}\Delta T$$
$$\therefore$$ $$\Delta T = {{M{v^2}\left( {\gamma - 1} \right)} \over {2R}}K$$