Question

One liter of a gas weighs $2g$ at $300K$ and $1atm$, at which of the following temperature, will one liter of the same gas weight $1g$ ?

• A. $450K$
• B. $600K$
• C. $800K$
• D. $900K$

Answer 1

Answer: (A)

$450K$

Given

$P₁=1atm;W₁=2gP₂=0.75atm;W₂=1gT₁=300K;V₁=1LT₂=?;V₂=1L$

we know that,

$\frac{P₁V₁}{P₂V₂}=\frac{W₁T₁}{W₂T₂}$

Substituting the values

$=>\frac{1\times 1}{0.75\times 1}=\frac{2\times 300}{1\times T₂}$

$=>\frac{1}{0.75}=\frac{600}{T₂}$

Cross Multiplying we get,

$=>T₂\times 1=600\times 0.75$

$=>T₂=450K$

Therefore the required temperature is 450 K