Question 565: Thermodynamics & KTG

Question

A gas of certain mass filled in a closed cylinder at a pressure of 3.23 kPa has temperature $50^{\circ} \mathrm{C}$. The gas is now heated to double its temperature. The modified pressure is $\_\_\_\_$ Pa .

Accepted Answer

The problem states the gas is filled in a closed cylinder. That means the volume of the gas remains constant during the heating process ( $\mathrm{V}=$ constant).
For a fixed mass of gas at constant volume, the pressure is directly proportional to the absolute temperature.
$$ \frac{\mathrm{P}_1}{\mathrm{~T}_1}=\frac{\mathrm{P}_2}{\mathrm{~T}_2} $$
Initial pressure is $\mathrm{P}_1=3.23 \mathrm{kPa}=3.23 	imes 1000 \mathrm{~Pa}=3230 \mathrm{~Pa}$
Initial temperature is $\mathrm{T}_1=50^{\circ} \mathrm{C}=(50+273) \mathrm{K}=323 \mathrm{~K}$
The final temperature is $\mathrm{T}_2=2 	imes \mathrm{T}_1=2 	imes 323 \mathrm{~K}=646 \mathrm{~K}$
Using the relation $\frac{\mathrm{P}_1}{\mathrm{~T}_1}=\frac{\mathrm{P}_2}{\mathrm{~T}_2}$ :
$$ \mathrm{P}_2=\mathrm{P}_1 	imes \frac{\mathrm{T}_2}{\mathrm{~T}_1} $$
$$\Rightarrow $$ $P_2=3230 	imes \frac{646}{323}$
$$ \mathrm{P}_2=3230 	imes 2=6460 \mathrm{~Pa} $$
Therefore, the modified pressure is 6460 Pa . Hence, the correct answer is 6460 .
Other Solution :
$\begin{aligned} & \mathrm{V}=	ext { constant } \ & 	ext { so } \mathrm{P} \propto \mathrm{T} \ & \mathrm{T}_{\mathrm{i}}=50^{\circ} \mathrm{C}=323 \mathrm{~K} \ & \mathrm{~T}_{\mathrm{f}}=100^{\circ} \mathrm{C}=373 \mathrm{~K} \ & \Rightarrow \frac{\mathrm{P}_{\mathrm{f}}}{\mathrm{P}_{\mathrm{i}}}=\frac{\mathrm{T}_{\mathrm{f}}}{\mathrm{T}_{\mathrm{i}}} \ & \Rightarrow \frac{\mathrm{P}_{\mathrm{f}}}{3.23 \mathrm{kPa}}=\frac{373}{323} \ & \Rightarrow \mathrm{P}_{\mathrm{f}}=3730 \mathrm{~Pa}\end{aligned}$

Thermodynamics & KTG

A gas of certain mass filled in a closed cylinder at a pressure of 3.23 kPa has temperature 50^{\circ} \mathrm{C}. The gas is now heated to double its temperature. The modified pressure is \_\_\_\_ Pa .

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