One litre of oxygen at a pressure of 1 atm, and 2 litres of nitrogen at a pressure of 0.5 atm. are introduced in the vessel of 1 litre capacity, without any change in temperature. The total pressure would be

1.5 atm.
0.5 atm.
2.0 atm.
1.0 atm.

Solution

A sample of an ideal gas occupies a volume of V at a pressure P and absolute temperature. T. The mass of each molecule is m. The equation for density is

m k T
P/k T
P/(k T V)
P m/k T

Solution

A vessel contains air at a temperature of 15º C and 60 % R.H.What will be the R.H. if it is heated to 20º C? (S.V.P. at 15º C is 12.67 & at 20º C is 17.36 mm of Hg respectively)

262 %
26.2 %
44.5 %
46.2 %

Solution

A gas mixture consists of molecules of type 1, 2 and 3, with molar masses m₁ > m₂ > m₃. v_rms and \(\bar{k}\) are the r.m.s. speed and average kinetic energy of the gases. Which of the following is true?

(v_rms)₁ < (v_rms)₂ < (v_rms)₃ and (\(\bar{k}\))₁ = (\(\bar{k}\))₂ = (\(\bar{k}\))₃
(v_rms)₁ > (v_rms)₂ > (v_rms)₃ and \(\bar{k}_{1} < \bar{k}_{2} > \bar{k}_{3}\)
(v_rms)₁ = (v_rms)₂ = (v_rms)₃ and (\(\bar{k}\))₁ = (\(\bar{k}\))₂ > (\(\bar{k}\))₃
(v_rms)₁ > (v_rms)₂ > (v_rms)₃ and \(\overline{k}_{1} < \overline{k}_{2} > \overline{k}_{3}\)

Solution

v_rms α ¹⁄_√M ⇒ (v_rms)₁ < (v_rms)₂ < (v_rms) also in mixture temperature of each gas will be same, hence kinetic energy also remains same.

If 2 mole of an ideal monatomic gas at temperature T₀ is mixed with 4 moles of another ideal monatomic gas at temperature 2 T₀, then the temperature of the mixture is

⁵⁄₃T₀
³⁄₂T₀
⁴⁄₃T₀
⁵⁄₄T₀

Solution

How many degrees of freedom are associated with 2 grams of He at NTP ?

3
3.01 × 10²³
9.03 × 10²³
6

Solution

Moles of He = ²⁄₄ = ¹⁄₂

Molecules = ¹⁄₂ × 6.02 × 10²³= 3.01 × 10²³

As there are 3 degrees of freedom corresponding to 1 molecule of a monatomic gas.

∴ Total degrees of freedom = 3 × 3.01 × 10²³

= 9.03 × 10²³

For a gas if ratio of specific heats at constant pressure and volume is γ then value of degrees of freedom is

\(\frac{3\gamma -1}{2\gamma -1}\)
\(\frac{2}{\gamma -1}\)
⁹⁄₂(γ - 1)
²⁵⁄₂(γ - 1)

Solution

\(\gamma =1+\frac{2}{f},\Rightarrow \gamma -1=\frac{2}{f}\Rightarrow \frac{f}{2}=\frac{1}{\gamma -1}\Rightarrow f=\frac{2}{\gamma -1}\)

Which of the following formula is wrong?

\(C_{v}=\frac{R}{\gamma -1}\)
\(C_{p}=\frac{\gamma R}{\gamma -1}\)
C_p/C_v = γ
C_p - C_v = 2R

Solution

The difference of C_p and C_v is equal to R, not 2R

One mole of a gas occupies 22.4 lit at N.T.P. Calculate the difference between two molar specific heats of the gas.J = 4200 J/kcal.

1.979 k cal/kmol K
2.378 k cal/kmol K
4.569 k cal/kmol K
3.028 k cal/ kmol K

Solution

Helium gas is filled in a closed vessel (having negligible thermal expansion coefficient) when it is heated from 300 K to 600 K, then average kinetic energy of helium atom will be

2 times
2 times
unchanged
half

Solution

The temperature of the mixture of one mole of helium and one mole of hydrogen is increased from 0°C to 100°C at constant pressure. The amount of heat delivered will be

600 cal
1200 cal
1800 cal
3600 cal

Solution

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