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 (integer)
60
Solution
We know that work done is <br><br>$W = \int {PdV}$ .... (1)<br><br>$\Rightarrow P = {{nRT} \over V}$ .... (2)<br><br>$\Rightarrow W = \int {{{nRT} \over V}dv}$ .... (3)<br><br>and given $V = K{T^{2/3}}$ .... (4)<br><br>$\Rightarrow W = \int {{{nRT} \over {K{T^{2/3}}}}.dv}$ .... (5)<br><br>$\Rightarrow$ from (4) : $dv = {2 \over 3}K{T^{ - 1/3}}dT$<br><br>$$ \Rightarrow W = \int\limits_{{T_1}}^{{T_2}} {{{nRT} \over {K{T^{2/3}}}}{2 \over 3}K{1 \over {{T^{1/3}}}}} dT$$<br><br>$\Rightarrow W = {2 \over 3}nR \times \left( {{T_2} - {T_1}} \right)$ .... (6)<br><br>$\Rightarrow {T_2} - {T_1} = 90K$ .... (7)<br><br>$\Rightarrow W = {2 \over 3}nR \times 90$<br><br>$\Rightarrow W = 60nR$<br><br>Assuming 1 mole of gas<br><br>n = 1<br><br>So, W = 60R
About this question
Subject: Physics · Chapter: Thermodynamics · Topic: Zeroth and First Law
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