The reciprocal of compressibility factor of a real gas in the critical state is

³⁄₈
³⁄₄
⁸⁄₃
¹⁄₃

Solution

When an ideal gas undergoes unrestrained expansion,no cooling occurs because the molecules

are above the inversion temperature
exert no attractive forces on each other
do work equal to loss in kinetic energy
collide without loss of energy

Solution

No work is required to tear apart the molecules due to the absence of attractive forces in an ideal gas.

The beans are cooked earlier in pressure cooker, because

boiling point increases with increasing pressure
boiling point decreases with increasing pressure
internal energy is not lost while cooking in pressure cooker
extra pressure of pressure cooker, softens the beans

Solution

The beans are cooked earlier in pressure cooker because boiling point increases with increasing pressure.

If 500 ml of gas A at 400 torr and 666.6 ml of B at 600 torr are placed in a 3 litre flask, the pressure of the system will be

200 torr
100 torr
550 torr
366 torr

Solution

14 g of N₂ and 36 g of ozone are at the same pressure and temperature . Their volumes will be related as

2V_N₂ = 3V_O₃
3V_N₂ = 2V_O₃
3V_N₂ = 4V_O₃
4V_N₂ = 3V_O₃

Solution

Cyclopropane and oxygen at partial pressures 170 torr and 570 torr respectively are mixed in a gas cylinder. What is the ratio of the number of moles of cyclopropane to the number of moles of oxygen (nC₃H₆/nO₂)?

\(\frac{170 \times 42}{570 \times 32} = 0.39\)
\(\frac{170}{42}/\left ( \frac{170}{42} + \frac{570}{32} \right ) \approx 0.19\)
\(\frac{170}{740} = 0.23\)
\(\frac{170}{570} = 0.30\)

Solution

By Ideal gas equation

P₁V = n₁RT

n₁ ∝ P₁ and n₂ ∝ P₂

^n₁⁄_n₂ = ^p₁⁄_p₂ ⇒ ^n₁⁄_n₂ = ¹⁷⁰⁄₅₇₀ = 0.30

The average kinetic energy of 28 g CO at,300 K is E kcal. The average kinetic energy of 2 g H₂ at the same temperature would be…..k cal.

E
14 E
1/14 E
28 E

Solution

28g of CO = ²⁸⁄₂₈ = 1 mole; 2g H₂ = ²⁄₂ = 1 mole

Since the temperature and moles are the same,the average kinetic energy will also be the same.

Under what conditions will a pure sample of an ideal gas not only exhibit a pressure of 1 atm but also a concentration of 1 mole litre^-1? (R = 0.082 litre atm mol^-1deg^-1)

At STP
When V= 22.4 litres
When T = 12 K
Impossible under any conditions

Solution

PV = nRT or P = ⁿ⁄_VRT = CRT.

Hence 1 = 1 × 0.082 × T is T = ¹⁄_0.082 = 12K

Reducing the pressure from 1.0 atm to 0.5 atm would change the number of molecules in one mole of ammonia to

25% of its initial value
50% of its initial value
75% of its initial value
None of the above

Solution

One mole of a substance contains the number of molecules which is independent of pressure.

The molecular velocities of two gases at the same temperature are u₁ and u₂. Their masses are m₁ and m₂ respectively. Which of the following expression is correct.

m₁/u₁² = m₂/u₂²
m₁u₁ = m₂u₂
m₁/u₁ = m₂/u₂
m₁u₁² = m₂u₂²

Solution

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