Question 95: Thermodynamics & KTG

Question

Two closed vessels of same volume are joined through a narrow tube and both vessels are filled with air of pressure 90 kPa and temperature 400 K . Keeping the temperature of one vessel constant at 400 K the second vessel temperature is raised to 500 K . The final pressure in the vessels is $\_\_\_\_$ kPa .

Accepted Answer

Using Ideal Gas Law:
PV = nRT
Where P is pressure, V is volume, n is number of moles, R is the gas constant, and T is absolute temperature. Since the system is closed, the total number of moles ( $\mathrm{n}_{	ext {total }}$ ) remains constant before and after heating.
Initial State:
Both vessels (each of volume V ) are at $\mathrm{P}_0=90 \mathrm{kPa}$ and $\mathrm{T}_0=400 \mathrm{~K}$.
$$ \mathrm{n}_{	ext {total }}=\frac{\mathrm{P}_0 \mathrm{~V}}{\mathrm{RT}_0}+\frac{\mathrm{P}_0 \mathrm{~V}}{\mathrm{RT}_0}=\frac{2 \mathrm{P}_0 \mathrm{~V}}{\mathrm{RT}_0} \ldots (i)$$
Final State: Pressure becomes $\mathrm{P}_{\mathrm{f}}$ in both vessels. Vessel 1 stays at $\mathrm{T}_1=400 \mathrm{~K}$, and vessel 2 is at $\mathrm{T}_2=500 \mathrm{~K}$.
$$ \mathrm{n}_{	ext {total }}=\frac{\mathrm{P}_{\mathrm{f}} \mathrm{~V}}{\mathrm{RT}_1}+\frac{\mathrm{P}_{\mathrm{f}} \mathrm{~V}}{\mathrm{RT}_2}=\frac{\mathrm{P}_{\mathrm{f}} \mathrm{~V}}{\mathrm{R}}\left(\frac{1}{\mathrm{~T}_1}+\frac{1}{\mathrm{~T}_2}ight) \ldots (ii)$$
From both the equations,
$$ \frac{2 \mathrm{P}_0 \mathrm{~V}}{\mathrm{RT}_0}=\frac{\mathrm{P}_{\mathrm{f}} \mathrm{~V}}{\mathrm{R}}\left(\frac{1}{\mathrm{~T}_1}+\frac{1}{\mathrm{~T}_2}ight) \Rightarrow \frac{2 \mathrm{P}_0}{\mathrm{~T}_0}=\mathrm{P}_{\mathrm{f}}\left(\frac{1}{\mathrm{~T}_1}+\frac{1}{\mathrm{~T}_2}ight) $$
Substituting the values ( $\mathrm{P}_0=90, \mathrm{~T}_0=400, \mathrm{~T}_1=400, \mathrm{~T}_2=500$ ):
$$ 2 	imes \frac{90}{400}=\mathrm{P}_{\mathrm{f}}\left(\frac{1}{400}+\frac{1}{500}ight) $$
$$\Rightarrow $$ $\frac{180}{400}=\mathrm{P}_{\mathrm{f}}\left(\frac{5+4}{2000}ight)$
$$\Rightarrow $$ $0.45=\mathrm{P}_{\mathrm{f}}\left(\frac{9}{2000}ight)$
$$\Rightarrow $$ $$ P_f=\frac{0.45 	imes 2000}{9}=\frac{900}{9}=100 \mathrm{kPa} $$
Therefore, the final pressure of the vessel is 100 kPa . Thus, the correct option is (A).

Answer

100

Answer

120

Answer

90

Answer

105

Thermodynamics & KTG

Choose the correct answer: