Question

1 litre of a gas weighs 2g at 300K and 1 atm pressure.If pressure is made 75atm at which temperature will 1 litre weigh 1g

Answer 1

Answer:-

45000 K

Explanation:-

We know that :-

n = w/M

Where :-

• n is the number of mole.

• w is given mass.

• M is molar mass.

∴ n₁ = w₁/M

∴ n₂ = w₂/M

Let the initial pressure, volume, temperature be P₁, V₁ and T₁ respectively. And let the final pressure, volume and temperature be P₂, V₂, and T₂.

Case 1 :-

According to Ideal Gas equation :-

=> P₁V₁ = n₁RT₁

=> 1×1 = w₁/M × 300R

=> 1 = 2/M × 300R

=> 1 = 600R/M -----(1)

Case (2) :-

According to Ideal Gas equation :-

=> P₂V₂ = n₂RT₂

=> 75×1 = w₂/M × RT₂

=> 75 = 1/M × RT₂

=> 75 = RT₂/M -----(2)

On dividing eq.2 by eq.1, we get :-

=> 75/1 = [600R/M]/[RT₂/M]

=> 75 = T₂/600

=> T₂ = 600(75)

=> T₂ = 45000 K

Thus, the required temperature is 45000K .

Answer 2

Explanation:

Answer:-

45000 K

Explanation:-

We know that :-

n = w/M

Where :-

• • n is the number of mole.
• • w is given mass.
• • M is molar mass.

∴ n₁ = w₁/M

∴ n₂ = w₂/M

• Let the initial pressure, volume, temperature be P₁, V₁ and T₁ respectively. And let the final pressure, volume and temperature be P₂, V₂, and T₂.

Case 1 :-

• According to Ideal Gas equation :-

=> P₁V₁ = n₁RT₁

=> 1×1 = w₁/M × 300R

=> 1 = 2/M × 300R

=> 1 = 600R/M -----(1)

Case (2) :-

• According to Ideal Gas equation :-

=> P₂V₂ = n₂RT₂

=> 75×1 = w₂/M × RT₂

=> 75 = 1/M × RT₂

=> 75 = RT₂/M -----(2)

• On dividing eq.2 by eq.1, we get :-

=> 75/1 = [600R/M]/[RT₂/M]

=> 75 = T₂/600

=> T₂ = 600(75)

=> T₂ = 45000 K

Thus, the required temperature is 45000K .

hope this helps you!!