Question

one liter of gas weighs 2g at 300k and 1atm of pressure is made 0.75 atm at which temperature will one litre of same ga weighs 1g

Answer 1

Hey !

Solution :

Assembling all the data we get,

P₁ = 1 atm ; W₁ = 2g

P₂ = 0.75 atm ; W₂ = 1 g

T₁ = 300 K ; V₁ = 1 L

T₂ = ? ; V₂ = 1 L

So we know that,

P₁V₁ / P₂V₂ = W₁T₁ / W₂T₂

Substituting in the above formula we get,

=> 1 × 1  / 0.75 × 1 = 2 × 300 / 1 × T₂

=>  1 / 0.75 = 600 / T₂

Cross Multiplying we get,

=> T₂ × 1 = 600 × 0.75

=> T₂ = 450 K

Therefore the required temperature is 450 K.

Hope it helped :-)

Answer 2

Solution :

Assembling all the data we get,

P₁ = 1 atm ; W₁ = 2g

P₂ = 0.75 atm ; W₂ = 1 g

T₁ = 300 K ; V₁ = 1 L

T₂ = ? ; V₂ = 1 L

So we know that,

P₁V₁ / P₂V₂ = W₁T₁ / W₂T₂

Substituting in the above formula we get,

=> 1 × 1 / 0.75 × 1 = 2 × 300 / 1 × T₂

=> 1 / 0.75 = 600 / T₂

Cross Multiplying we get,

=> T₂ × 1 = 600 × 0.75

=> T₂ = 450 K

Therefore the required temperature is 450 K.

Hope it helped :-)