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
Answers
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 .
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 .