Question

2 gm of monoatomic gas occupies 2 litre volume at 8.3 × 10∧5 pascal and at 127°C, then the gas is 1] Hydrogen 2] Neon 3] Helium 4] Argon

Answer 1

Answer:Helium

Explanation:PV=m/M×Rt

2×10^-3×8.3×0^5

=2×10^3/M×8.3×400

M=2×10^3×8.3×400/2×10^-3×8.3×10^5

M=4 gram/mole

HELIUM

Answer 2

Answer:

Monoatomic gas is Helium. The molecular mass of Helium is equal to 4gmol⁻¹.

Therefore the option (3) is correct.

Explanation:

From the ideal gas equation:

$PV=nRT$

We know, number of moles $n=\frac{m}{M}$

Therefore, $PV=\frac{m}{M}RT$                                            ....................(1)

Given: The mass of monoatomic gas, $m=2g=2*10^{-3}kg$

Volume of gas, $V=2L=3*10^{-3}m^{3}$

Temperature, $T=273+127=400K$

Pressure of monoatomic gas, $P=8.3*10^{5}Pa=8.3*10^{5}Kgm^{-1}s^{-2}$

The value of Gas constant, $R=8.314JK^{-1}mol^{-1}$

Put all the values in eq.(1);

$(8.3*10^{5}kgm^{-1}s^{-2})(2*10^{-3}m^{3})=(\frac{2*10^{-3}kg}{M} )(8.314JK^{-1}mol^{-1})(400K)$

$M(16.6*10^{2}m^{2}s^{-2})=6651.2*10^{-3}kgm^{2}s^{-2}mol^{-1}$

where 1Pa = 1kg m⁻¹s²  and 1Jolue = kgm²s⁻²

$M=400*10^{-5}Kgmol^{-1}\\M=4*10^{-3}*10^{3}gmol^{-1}\\M=4gmol^{-1}$

The molecular mass of monoatomic gas is 4gmol⁻¹. Therefore, the given monoatomic gas is Helium.