At what centigrade temperature will the volume of gas becomes 2x, if the volume of this gas is ‘x’ at 0° C at constant pressure?

0°C
237°C
273°C
546°C

Solution

By ideal gas equation

\(\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\)

Given that,

P₁ = P₂, V₁ = x, V₂ = 2x,

T₁ = 273 K, T₂ = ?

On putting value

\(\frac{P_{2}x}{273} = \frac{P_{2}2x}{T_{2}}\)

T₂ = 2 × 273 = 546 K or 273°C

Hence, option (d) is correct.

In van der Waal’s equation of state of the gas law, the constant ‘b’ is a measure of

volume occupied by the molecules
inter molecular attraction
inter molecular repulsions
inter molecular collisions per unit volume

Solution

In vander waals equation ‘b’ is for volume correction

At which one of the following temperature-pressure conditions the deviation of a gas from ideal behaviour is expected to be minimum?

350 K and 3 atm.
550 K and 1 atm.
250 K and 4 atm.
450 K and 2 atm.

Solution

At low pressure and high temperature real gas nearly behave like ideal gas. Hence deviation is minimum from ideal behaviour.

Some moles of O₂ diffuse through a small opening in 18 s. Same number of moles of an unknown gas diffuse through the same opening in 45 s. molecular weight of the unknown gas is :

\(32 \times \frac{45^{2}}{18^{2}}\)
\(32 \times \frac{18^{2}}{45^{2}}\)
\(32 \times \frac{45}{18}\)
\(32^{2} \times \frac{18^{2}}{45^{2}}\)

Solution

When is deviation more in the behaviour of a gas from the ideal gas equation PV = nRT ?

At high temperature and low pressure
At low temperature and high pressure
At high temperature and high pressure
At low temperature and low pressure

Solution

At low temperature and high pressure.

An ideal gas can’t be liquefied because

its critical temperature is always above 0°C
Its molecules are relatively smaller in size
it solidifies before becoming a liquid
forces between its molecules are negligible

Solution

In the ideal gas, the inter molecular forces of attraction are negligible and hence it cannot be liquefied.

The total pressure of a mixture of two gases is:

The sum of the partial pressures
The difference between the partial pressures
The product of the partial pressures
The ratio of the partial pressures.

Solution

By Dalton’s law of partial pressures, the total pressure of a mixture of two gases is the sum of the partial pressures.

The root mean square velocity of one mole of a monoatomic gas having molar mass M is u_r.m.s.. The relation between the average kinetic energy (E) of the gas and u_r.m.s. is

\(u_{r.m.s.} = \sqrt{\frac{3E}{2M}}\)
\(u_{r.m.s.} = \sqrt{\frac{2E}{3M}}\)
\(u_{r.m.s.} = \sqrt{\frac{2E}{M}}\)
\(u_{r.m.s.} = \sqrt{\frac{E}{3M}}\)

Solution

Average KE = E = ¹⁄₂Mu²_rms

∴ u²_rms = ^2E⁄_M or \(u_{r.m.s.} = \sqrt{\frac{2E}{M}}\)

The root mean square velocity of an ideal gas at constant pressure varies with density (d) as

d²
d
√d
1/√d

Solution

The ratio between the root mean square speed of H₂ at 50 K and that of O₂ at 800 K is,

Solution

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