Question

The weight of one lite ol a gas ar laim. Pressure and 300Kg. Alhamtare
WAS IS 4g when the pressure is made O Sam and volume is I live?
I) 200K
2) 150K
3) 600K
4) 1200K​

Answer 1

Answer:

Explanation:

P₁ = 1 atm ; W₁ = 4g

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 = 600 K

Therefore the required temperature is 600 K.

Answer 2

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Refer the attachment.