JEE MAIN - Physics (2022 - 26th June Morning Shift - No. 8)
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)
$${{M{v^2}} \over {7R}}$$
$${{M{v^2}} \over {5R}}$$
2$${{M{v^2}} \over {7R}}$$
7$${{M{v^2}} \over {5R}}$$
Explanation
Let there be n moles of gas
Eloss = Egain
$${1 \over 2}(nM){v^2} = n{C_v}\Delta T$$
$${1 \over 2}M{v^2} = {C_v}\Delta T$$
here, $$\gamma = 1.4 = {7 \over 5}$$ i.e. diatomic gas
$$\therefore$$ $${C_v} = {{5R} \over 2}$$
Now, $${1 \over 2}M{v^2} = {{5R} \over 2}\Delta T$$
$$\Delta T = {{M{v^2}} \over {5R}}$$
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