Question 91: Thermodynamics & KTG

Question

One mole of diatomic gas having rotational modes only is kept in a cylinder with a piston system. The cross-section area of the cylinder is $4 \mathrm{~cm}^2$. The gas is heated slowly to raise the temperature by $1.2^{\circ} \mathrm{C}$ during which the piston moves by 25 mm . The amount of heat supplied to the gas is $\_\_\_\_$ J.
(Atmospheric pressure $=100 \mathrm{kPa}, R=8.3 \mathrm{~J} / \mathrm{mol} . \mathrm{K}$ ) (Neglect mass of the piston)

Accepted Answer

To find the total heat supplied ( Q ), we use the $1^{	ext {st }}$ law of thermodynamics :
$$ Q=\Delta U+W $$
Where, $\Delta \mathrm{U}$ is the change in internal energy and W is the work done by the gas.
The internal energy of an ideal gas depends on its degrees of freedom (f). For a diatomic gas, there are 3 translational and 2 rotational degrees of freedom ( $f=5$ ). However, it is given that the gas has rotational modes only, which means we only consider $f=2$.
The formula for internal energy change is :
$$ \Delta U=\frac{n \mathrm{fR} \Delta T}{2} $$
The amount of gas is 1 mole, $\mathrm{n}=1$ mole, the degree of freedom is $\mathrm{f}=2$, the universal gas constant is $\mathrm{R}= \frac{8.3 \mathrm{~J}}{\mathrm{~mol} \cdot \mathrm{~K}}$ and the change in temperature of gas is $\Delta \mathrm{T}=1.2^{\circ} \mathrm{C}=1.2 \mathrm{~K}$
$$ \Delta \mathrm{U}=\frac{1 	imes 2 	imes 8.3 	imes 1.2}{2} $$
$$\Rightarrow $$ $$ \Delta U=8.3 	imes 1.2=9.96 \mathrm{~J} $$
Work done by the gas in a piston system is given by:
$$ \mathrm{W}=\mathrm{P} \Delta \mathrm{~V} $$
Since the mass of the piston is neglected and it moves slowly, the internal pressure P equals the atmospheric pressure ( $\mathrm{P}_{\mathrm{atm}}$ ).
The change in volume $\Delta \mathrm{V}$ is $\Delta \mathrm{V}=\mathrm{A} 	imes \Delta \mathrm{x}$, where the cross-sectional area is A and displacement of the piston is $\Delta \mathrm{x}$.
The atmospheric pressure is $\mathrm{P}=100 \mathrm{kPa}=10^5 \mathrm{~Pa}$
The area of cross-section is $\mathrm{A}=4 \mathrm{~cm}^2=4 	imes 10^{-4} \mathrm{~m}^2$
The displacement of piston $\mathrm{d}=25 \mathrm{~mm}=25 	imes 10^{-3} \mathrm{~m}$
$$ \Delta V=A 	imes \Delta x=\left(4 	imes 10^{-4}ight) 	imes\left(25 	imes 10^{-3}ight)=100 	imes 10^{-7}=10^{-5} \mathrm{~m}^3 $$
So, the work done by the gas is;
$$ \mathrm{W}=10^5 	imes 10^{-5}=1 \mathrm{~J} $$
Using $1^{	ext {st }}$ law of thermodynamics,
$$ \mathrm{Q}=\Delta \mathrm{U}+\mathrm{W} $$
$$\Rightarrow $$ $$ Q=9.96 \mathrm{~J}+1 \mathrm{~J}=10.96 \mathrm{~J} $$
Therefore, the amount of heat supplied by the gas is 10.96 J.

Answer

24.8

Answer

10.96

Answer

15.04

Answer

29.98

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