Pressure and Temperature
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It is inverted into a tray containing mercury and the cork is opened. The height of mercury column now in the tube over the surface of mercury in the tray will be:
- 76 cm
- Zero
- >76 cm
- <76 cm
- 109 K
- 273 K
- 373 K
- 0 K
Why solids are rigid?
- 134.67 kPa
- 156.37 kPa
- 143.32 kPa
- 75.75 kPa
- 136.42 kPa
- 65.98 kPa
- 101.25 kPa
- 91.34 kPa
Given: Density of water = 1000 kgm3, Density of mercury = 13600 kgm3
- 0.76 m
- 10.3 m
- 11.2 m
- 9.8 m
- Internal energy of the gas remains unchanged
- Internal energy of the gas decrease
- Temperature of the system decreases
- ΔE+PΔV=0
Solids have the smallest inter-molecular space and the strongest inter-molecular force of attraction among its molecules.
- True
- False
दी गयी दो पदीय ऊष्माशोषी अभिक्रिया के लिए निम्नलिखित में से कौनसा ग्राफ सही है?
Why solids are hard and rigid?
- H
- 2H
- 7H
- 8H
Which of the following statement is correct about solids?
They are less rigid.
They show maximum expansion on heating.
They have large intermolecular distance between the molecules.
They have the strongest intermolecular force of attraction between the molecules.
Why can the molecules of solids not interchange their positions?
- Complete order of molecules
- Complete diorder of molecules
- Random motion of molecules
- Fixed position of molecules
The critical temperature of gas and thegas are 31oC and -1470C, respectively. Which gas is liquefied easily and why?
Given: Density of mercury = 13600 kg/m3
- 546.8 cm
- 742.6 cm
- 432.5 cm
- 861.3 cm
Identify the wrong statement among the following.
In solids, molecules are closely packed.
In solids, molecules have very strong intermolecular forces of attraction.
In solids, molecules have only vibratory motion about their mean positions.
In solids, molecules possess very high kinetic energy
- 10.5 g/cm3
- 4.8 g/cm3
- 3.4 g/cm3
- 5.6 g/cm3
- 5040 kg
- 4340 kg
- 3500 kg
- 5740 kg
Two spheres - 1, 2 with flexible walls are connected by a tube of negligible volume. Their initial and final conditions are given by P1, V1, n1 and P2, V2, n2 respectively. The radius of these spheres depends on the temperature as 4πr2dr=kdT. Initially, sphere 1 was at 100 K and sphere 2 was at 200 K. As the temperature of sphere 1 is increased to 300 K, find the final mole ratio (n′1n′2).