The amount of work done in Joule to increase the temperature of one mole of ideal gas by 300 C if it is expanding under the condition V - T 23 is R - 8-31 J - mol - K

V = kT^2/3 dV = 2/3 kT^ 1/3 dT , W = ∫ PdV = ∫nRTdV/V = R ∫T/V dV = nbsp; 2/3 R ∫(TKT^ 1/3dT)/(KT^2/3) = 2/3 R (T_2 T_1) = 2/3 R (30) = 20 (8.31) = 166.2 J ...

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Standard XII

Physics

2nd Law of Thermodynamics

The amount of...

Question

The amount of work done (in Joule) to increase the temperature of one mole of ideal gas by $30^{o} C$ if it is expanding under the condition $V \propto T^{\frac{2}{3}}$ is (R = 8.31 J/mol-K)

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Solution

$V = k T^{\frac{2}{3}}$
$d V = \frac{2}{3} k T^{\frac{- 1}{3}} d T, W = \int P d V = \int \frac{n R T d V}{V}$
$= R \int \frac{T}{V} d V = \frac{2}{3} R \int \frac{(T K T^{\frac{- 1}{3}} d T)}{(K T^{\frac{2}{3}})}$
$= \frac{2}{3} R (T_{2} - T_{1}) = \frac{2}{3} R (30) = 20 (8.31) = 166.2 J$

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The amount of work done (in Joule) to increase the temperature of one mole of ideal gas by 30oC if it is expanding under the condition V∝T23 is (R = 8.31 J/mol-K)

V=kT23 dV=23kT−13dT,W=∫PdV=∫nRTdVV =R∫TVdV=23R∫(TKT−13dT)(KT23) =23R(T2−T1)=23R(30)=20(8.31)=166.2J

The amount of work done (in Joule) to increase the temperature of one mole of ideal gas by $30^{o} C$ if it is expanding under the condition $V \propto T^{\frac{2}{3}}$ is (R = 8.31 J/mol-K)

$V = k T^{\frac{2}{3}}$
$d V = \frac{2}{3} k T^{\frac{- 1}{3}} d T, W = \int P d V = \int \frac{n R T d V}{V}$
$= R \int \frac{T}{V} d V = \frac{2}{3} R \int \frac{(T K T^{\frac{- 1}{3}} d T)}{(K T^{\frac{2}{3}})}$
$= \frac{2}{3} R (T_{2} - T_{1}) = \frac{2}{3} R (30) = 20 (8.31) = 166.2 J$