JEE MAIN - Physics (2021 - 25th February Morning Shift - No. 20)
A monoatomic gas of mass 4.0 u is kept in an insulated container. Container is moving with velocity 30 m/s. If container is suddenly stopped then change in temperature of the gas (R = gas constant) is $${x \over {3R}}$$. Value of x is ___________.
Answer
3600
Explanation
$$\Delta {K_E} = \Delta U$$
$$\Delta U = n{C_v}\Delta T$$
$${1 \over 2}m{v^2} = {3 \over 2}nR\Delta T$$
$${{m{v^2}} \over {3nR}} = \Delta T$$
$${{4 \times {{(30)}^2}} \over {3 \times 1 \times R}} = \Delta T$$
$$ \Rightarrow $$ $$\Delta T = {{1200} \over R}$$
$$ \Rightarrow $$ $${x \over {3R}} = {{1200} \over R}$$
$$ \Rightarrow $$ $$x = 3600$$
$$\Delta U = n{C_v}\Delta T$$
$${1 \over 2}m{v^2} = {3 \over 2}nR\Delta T$$
$${{m{v^2}} \over {3nR}} = \Delta T$$
$${{4 \times {{(30)}^2}} \over {3 \times 1 \times R}} = \Delta T$$
$$ \Rightarrow $$ $$\Delta T = {{1200} \over R}$$
$$ \Rightarrow $$ $${x \over {3R}} = {{1200} \over R}$$
$$ \Rightarrow $$ $$x = 3600$$
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