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1. ⇒ (NEET 2023 Manipur )

A container of volume $200 {cm}^{3}$ contains 0.2 mole of hydrogen gas and 0.3 mole of argon gas. The pressure of the system at temperature $200 K$ ( $R = 8.3 {JK}^{- 1} {mol}^{- 1}$ ) will be :-

(A) $6.15 \times 10^{5} Pa$

(B) $6.15 \times 10^{4} Pa$

(C) $4.15 \times 10^{5} Pa$

(D) $4.15 \times 10^{6} Pa$

Correct answer is (D)

$\begin{aligned} P_{mix} = \frac{(μ_{1} + μ_{2}) {RT}_{mix}}{V_{mix}} \\ = \frac{(0.2 + 0.3) \times 8.3 \times 200}{200 \times 10^{- 6}} \\ = \frac{0.5 \times 8.3 \times 200}{200 \times 10^{- 6}} \\ P_{mix} = 4.15 \times 10^{6} P_{a} \end{aligned}$

2. ⇒ (NEET 2022 Phase 1 )

The volume occupied by the molecules contained in 4.5 kg water at STP, if the intermolecular forces vanish away is

(A) 5.6 $\times$ 10⁶ m³

(B) 5.6 $\times$ 10³ m³

(C) 5.6 $\times$ 10^{$-$ 3} m³

(D) 5.6 m³

Correct answer is (D)

From ideal gas equation

$P V = n R T$

$[n = \frac{m a s s o f w a t e r}{m o l . w t .} = \frac{4.5 \times 10^{3}}{18}]$

$V = \frac{n R T}{P}$

At $S T P \Rightarrow T = 273 K$

$P = 10^{5}$ N/m²

$V = \frac{4.5 \times 10^{3}}{18} \times \frac{8.3 \times 273}{10^{5}} = 5.66$ m³

3. ⇒ (NEET 2020 Phase 1 )

A cylinder contains hydrogen gas at pressure 249 kPa and temperature 27 $^{\circ}$ C
Its density is : (R = 8.3 J mol^-1 K^-1)

(A) 0.2 kg/m³

(B) 0.1 kg/m³

(C) 0.02 kg/m³

(D) 0.5 kg/m³

Correct answer is (A)

From the ideal gas equation, PV = nRT also
Volume (V) = $\frac{m a s s (M)}{d e n s i t y (ρ)}$
So, PM = $ρ$ RT
P = 249 $\times$ 10³ N/m²
M = 2 $\times$ 10^-3 kg
T = 333 K
$ρ$ = $\frac{(249 \times 10^{3}) (2 \times 10^{- 3})}{8.3 \times 300}$
= 0.2 kg/m³

4. ⇒ (NEET 2016 Phase 2 )

A given sample of an ideal gas occupies a volume V at a pressure P and absolute temperature T. The mass of each molecule of the gas is m. Which of the following gives the density of the gas ?

(A) P/(kT)

(B) Pm/(kT)

(C) P/(KTV)

(D) mkT

Correct answer is (B)

As PV = nRT

$\Rightarrow n = \frac{P V}{R T} = \frac{m a s s}{m o l a r m a s s}$    ...(i)

Density, $ρ$ = $\frac{m a s s}{v o l u m e} = \frac{(m o l a r m a s s) P}{R T} = \frac{(m N_{A}) P}{R T}$
   [From eqn. (i)]

$ρ = \frac{m P}{k T}$     $(∵ R = N_{A} k)$

5. ⇒ (AIPMT 2015 )

Two vessels separately contain two ideal gases A and B at the same temperature, the pressure of A being twice that of B. Under such conditions, the density of A is found to be 1.5 times the density of B. The ratio of molecular weight of A and B is

(A) 2

(B) $\frac{1}{2}$

(C) $\frac{2}{3}$

(D) $\frac{3}{4}$

Correct answer is (D)

From PV = nRT

$P_{A} = \frac{ρ_{A} M_{A}}{R T}$ and $P_{B} = \frac{ρ_{B} M_{B}}{R T}$

From question,

$\frac{P_{A}}{P_{B}} = \frac{ρ_{A}}{ρ_{B}} \frac{M_{A}}{M_{B}} = 2 \frac{M_{A}}{M_{B}} = \frac{3}{2}$

$∴$ $\frac{M_{A}}{M_{B}} = \frac{3}{4}$

6. ⇒ (NEET 2013 )

In the given (V $-$ T) diagram, what is the relation between pressure P₁ and P₂?

NEET 2013 Physics - Heat and Thermodynamics Question 55 English

(A) P₂ < P₁

(B) Carnot be predicted

(C) P₂ = P₁

(D) P₂ > P₁

Correct answer is (A)

NEET 2013 Physics - Heat and Thermodynamics Question 55 English Explanation
According to ideal gas equation PV = nRT

$\Rightarrow V = \frac{n R T}{P}$

For an isobaric process, P = constant and V $\propto$ T
Therefore, V – T graph is a straight line passing through origin. Slope of this line is inversely proportional to P.
In the given figure,

(Slope)₂ > (Slope)₁ $∴$ P₂ < P₁

7. ⇒ (AIPMT 2008 )

At 10^oC the value of the density of a fixed mass of an ideal gas divided by it pressure is x. At 110^oC this ratio is

(A) $\frac{10}{110} x$

(B) $\frac{283}{383} x$

(C) $x$

(D) $\frac{383}{283} x$

Correct answer is (B)

PV = nRT

$\Rightarrow P V = \frac{m}{M} R T$

$\Rightarrow \frac{P V}{m} = \frac{R T}{M}$

$\Rightarrow \frac{P}{ρ} = \frac{R T}{M}$

$\Rightarrow \frac{P}{ρ} \propto T$

$∴ \frac{P}{ρ} \propto \frac{1}{T}$

$\frac{\frac{P_{1}}{ρ_{1}}}{\frac{P_{2}}{ρ_{2}}} = \frac{T_{1}}{T_{2}} = \frac{383}{283}$

$\Rightarrow \frac{x}{\frac{P_{2}}{P_{2}}} = \frac{383}{283}$

$∴ \frac{P_{2}}{P_{2}} = \frac{383}{283} x$

8. ⇒ (AIPMT 2004 )

The equation of state for 5 g of oxygen at a pressure P and temperature T, when occupying a volume V, will be
(where R is the gas constant)

(A) PV = (5/32)RT

(B) PV = 5RT

(C) PV = (5/2)RT

(D) PV = (5/16) RT

Correct answer is (A)

As PV = nRT

$n = \frac{m}{m o l e c u l a r m a s s} = \frac{5}{32}$

$\Rightarrow P V = (\frac{5}{32}) R T$

⇒Kinetic Theory of Gases

Topic 01: Gas Law