KTGMedium
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 .
Options
A.100
B.120
C.90
D.105
Solution
Sol. [Key Concept: Conserve moles: $\frac{2P_{0}V}{{RT}_{0}} = \frac{PV}{{RT}_{0}} + \frac{PV}{{RT}_{1}}$ with $T_{0} = 400,\ T_{1} = 500$ K; solve for the final $P$.]
Using mole conservation
$${\frac{2 \times 9 \times 10^{4} \times V}{R \times 400} = \frac{P \times V}{R \times 400} + \frac{P \times V}{R \times 500} }{P = 100kPa }$$
Create a free account to view solution
View Solution FreeMore KTG Questions
Two containers of same volume are filled with atomic Hydrogen and Helium respectively at 1 and 2 atm pressure. If the te...The constant quantity for equal volume of two gases at same pressure and temperature is-...A mixture of 2 moles of helium gas (atomic mass = 4 amu) and 1 mole of argon gas (atomic mass = 40amu) is kept at 300 K ...Ratio of molecular specific heats of a di- atomic gas will be :-...The quantity represents (where U = internal energy of gas)...