JEE MAIN - Physics (2021 - 26th February Evening Shift - No. 24)
The volume V of a given mass of monoatomic gas changes with temperature T according to the relation $$V = K{T^{{2 \over 3}}}$$. The workdone when temperature changes by 90K will be xR. The value of x is _________. [R = universal gas constant]
Answer
60
Explanation
We know that work done is
$$W = \int {PdV} $$ .... (1)
$$ \Rightarrow P = {{nRT} \over V}$$ .... (2)
$$ \Rightarrow W = \int {{{nRT} \over V}dv} $$ .... (3)
and given $$V = K{T^{2/3}}$$ .... (4)
$$ \Rightarrow W = \int {{{nRT} \over {K{T^{2/3}}}}.dv} $$ .... (5)
$$ \Rightarrow $$ from (4) : $$dv = {2 \over 3}K{T^{ - 1/3}}dT$$
$$ \Rightarrow W = \int\limits_{{T_1}}^{{T_2}} {{{nRT} \over {K{T^{2/3}}}}{2 \over 3}K{1 \over {{T^{1/3}}}}} dT$$
$$ \Rightarrow W = {2 \over 3}nR \times \left( {{T_2} - {T_1}} \right)$$ .... (6)
$$ \Rightarrow {T_2} - {T_1} = 90K$$ .... (7)
$$ \Rightarrow W = {2 \over 3}nR \times 90$$
$$ \Rightarrow W = 60nR$$
Assuming 1 mole of gas
n = 1
So, W = 60R
$$W = \int {PdV} $$ .... (1)
$$ \Rightarrow P = {{nRT} \over V}$$ .... (2)
$$ \Rightarrow W = \int {{{nRT} \over V}dv} $$ .... (3)
and given $$V = K{T^{2/3}}$$ .... (4)
$$ \Rightarrow W = \int {{{nRT} \over {K{T^{2/3}}}}.dv} $$ .... (5)
$$ \Rightarrow $$ from (4) : $$dv = {2 \over 3}K{T^{ - 1/3}}dT$$
$$ \Rightarrow W = \int\limits_{{T_1}}^{{T_2}} {{{nRT} \over {K{T^{2/3}}}}{2 \over 3}K{1 \over {{T^{1/3}}}}} dT$$
$$ \Rightarrow W = {2 \over 3}nR \times \left( {{T_2} - {T_1}} \right)$$ .... (6)
$$ \Rightarrow {T_2} - {T_1} = 90K$$ .... (7)
$$ \Rightarrow W = {2 \over 3}nR \times 90$$
$$ \Rightarrow W = 60nR$$
Assuming 1 mole of gas
n = 1
So, W = 60R
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