Question 170: Thermodynamics & KTG

Question

The internal energy of a monoatomic gas is 3nRT. One mole of helium is kept in a cylinder having internal cross section area of $17 \mathrm{~cm}^2$ and fitted with a light movable frictionless piston. The gas is heated slowly by suppling 126 J heat. If the temperature rises by $4^{\circ} \mathrm{C}$, then the piston will move $\_\_\_\_$ cm.
(atmospheric pressure $=10^5 \mathrm{~Pa}$ )

Accepted Answer

According to the first law of thermodynamics,
$$ \Delta Q=\Delta U+\Delta W $$
Where,
The heat supplied to a system $=\Delta \mathrm{Q}=126 \mathrm{~J}$
The change in internal energy $=\Delta U=3 n R T$
The work done by the system $=\Delta \mathrm{W}$
Number of moles $n=1$ mole of Helium.
Change in temperature $=\Delta \mathrm{T}=4^{\circ} \mathrm{C}=4 \mathrm{~K}$
Since the piston is light, movable, and frictionless, the gas pressure remains equal to the atmospheric pressure $= 10^5 \mathrm{~Pa}$.
Area $=\mathrm{A}=17 \mathrm{~cm}^2=17 	imes 10^{-4} \mathrm{~m}^2$
Using the formula provided :
$$ \Delta \mathrm{U}=3 \cdot \mathrm{n} \cdot \mathrm{R} \cdot \Delta \mathrm{~T} $$
$$\Rightarrow $$ $\Delta \mathrm{U}=3 \cdot 1 \cdot 8.314 \cdot 4 \mathrm{~J}$
$$ \Delta \mathrm{U}=99.768 \mathrm{~J} $$
From the first law :
$$ \Delta \mathrm{W}=\Delta \mathrm{Q}-\Delta \mathrm{U} $$
$$\Rightarrow $$ $\Delta \mathrm{W}=126-99.768=26.232 \mathrm{~J}$
Work done by a gas at constant pressure is :
$$ \Delta \mathrm{W}=\mathrm{P} \Delta \mathrm{~V}=\mathrm{P} 	imes(\mathrm{Ax}) $$
Where x is the distance the piston moves.
$$ 26.232=10^5 	imes\left(17 	imes 10^{-4}ight) 	imes \mathrm{x} $$
$$\Rightarrow $$ $26.232=170 	imes x$
$$\Rightarrow $$ $x=\frac{26.232}{170} \approx 0.1543 \mathrm{~m}$
$$\Rightarrow $$ $\mathrm{x}=0.1543 	imes 100 \mathrm{~cm} \approx 15.5 \mathrm{~cm}$

Answer

1.55

Answer

14.5

Answer

15.5

Answer

1.45

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