Select one correct statement. In the gas equation, PV = nRT

n is the number of molecules of a gas
V denotes volume of one mole of the gas
n moles of the gas have a volume V
P is the pressure of the gas when only one mole of gas is present.

Solution

In the equation PV = nRT, n moles of the gas have volume V.

Under what conditions will a pure sample of ideal gas not only exhibit a pressure of 1 atm, but also a concentration of 2 moles per litre?

At STP
When volume is 22.41 litre
At 6.1 K
When R has no unit

Solution

Given P = 1 atm, n = 2 mole, V = 1L

Let at T K the given sample will exhibit pressure of 1 atm and a concentration of 2M.

For an ideal gas, PV = nRT

⇒ (1 atm) (1 L)

= (2 mol) (0.0821 mol^-1 L atm K^-1)T

⇒ T = \(\frac{1}{2\left( 0.0821 \right)}K = 6.0901 K \approx 6.1 K\)

So, T = 6.1 K is the required condition.

In two separate bulbs containing ideal gases A and B respectively, the density of gas A is twice of that of gas B, while mol wt.of gas A is half of that of gas B at the same temperature, pressure. PA/PB will be:

Solution

For an ideal gas, PV = nRT

⇒ \(P = \frac{n}{V}RT = \frac{mRT}{MV} = \frac{dRT}{M}\left ( \frac{m}{V} = density\left ( say \;\, d \right ) \right )\)

So, \(\frac{P_{A}}{P_{B}} = \frac{d_{A}M_{B}}{d_{B}M_{A}} = \frac{\left ( 2d_{B} \right ) \times M_{B}}{d_{B}\left ( M_{B}/2 \right )} = 4\)

The correct value of the gas constant ‘R’ is close to :

0.082 litre-atmosphere K
0.082 litre-atmosphere K^-1 mol^-1
0.082 litre– atmosphere^-1 K mol^-1
0.082 litre^-1 atmosphere^-1 K mol

Solution

R = 0.082 litre atm K^-1 mole^-1 .

Which is not true in case of an ideal gas ?

It cannot be converted into a liquid
There is no interaction between the molecules
All molecules of the gas move with same speed
At a given temperature, PV is proportional to the amount of the gas

Solution

Molecules in an ideal gas move with different speeds.

The molecular weight of two gases are 100 and 81 respectively. Their rates of diffusions are in the ratio:

81 : 100
100 : 81
10 : 9
9 : 10

Solution

According to Graham’s law of diffusion,

\(r \propto \sqrt{\frac{1}{M}}\)

where r is rate of diffusion of gas and M is its molcular weight

So, \(\frac{r_{1}}{r_{2}} = \sqrt{\frac{M_{2}}{M_{1}}} \Rightarrow \frac{r_{1}}{r_{2}} = \sqrt{\frac{81}{100}} = \frac{9}{10}\)

⇒ r₁ : r₂ = 9 : 10

At constant temperature, for a given mass of an ideal gas

the ratio of pressure and volume always remains constant.
volume always remains constant.
pressure always remains constant.
the product of pressure and volume always remains constant.

Solution

According to Boyle’s law at constant temperature,

V ∝ ¹⁄_P or PV = constant

The K.E. for 14 g of nitrogen gas at 127°C is nearly (molecular mass of nitrogen is 28 g/mole) and gas constant is 8.31 J/molK)

8.3 kJ
4.15 kJ
2.5 kJ
3.3 kJ

Solution

K. E of n moles of N₂ gas = ³⁄₂nRT

Here n = ¹⁴⁄₂₈ = ¹⁄₂ moles

R = 8.31 J/mol/K

T = 127°C = 400K

∴ KE = ³⁄₂ × ¹⁄₂ × 8.31 × 400J

= 2493.0 J = 2.493 kJ ≅ 2.5 kJ

A gas of volume of 15 ml at 300 K and 740 mm of Hg. Find the temperature if volume becomes 10 ml at 760 mm pressure of Hg.

209 K
200 K
205 K
275 K

Solution

Given: P₁ = 740 mm of Hg; V₁ = 15 mL; T₁ = 300 K and P₂ = 760 mm of Hg; V₂ = 10 mL; T₂ = ?

Using gas equation,

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

⇒ \(\frac{740 \times 15}{300} = \frac{760 \times 10}{T_{2}}\)

⇒ T₂ = \(\frac{760 \times 10 \times 300}{740 \times 15}\) = 205.405 ≅ 205 K

Correct gas equation is :

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

Solution

⇒ \(\frac{PV}{T}\) = constant or \(\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\)

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

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