Question

Given a $22.4$ liters sample of helium gas at STP, if the temperature is increased by 15 ${}^{o}C$ and the pressure is changed to $600$ torr, what would the new volume (in litres) ?

• A. $\frac{\left(760\right)\times \left(22.4\right)\times \left(15\right)}{\left(273\right)\times \left(600\right)}$
• B. $\frac{\left(273\right)\times \left(600\right)\times \left(288\right)}{\left(760\right)\times \left(22.4\right)}$
• C. $\frac{\left(760\right)\times \left(22.4\right)\times \left(15\right)}{\left(600\right)}$
• D. $\frac{\left(760\right)\times \left(22.4\right)\times \left(288\right)}{\left(273\right)\times \left(600\right)}$
• E. $\frac{\left(273\right)\times \left(600\right)}{\left(760\right)\times \left(22.4\right)\times \left(288\right)}$

Answer 1

Answer: (D)

$\frac{\left(760\right)\times \left(22.4\right)\times \left(288\right)}{\left(273\right)\times \left(600\right)}$

$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$
$P_1=1$ atm $=760$ torr
$V_1=22.4$ L
$T_1=273$ K
$P_2=600$ torr
$V_2=??$

$T_2=273+15=288$ K

$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

$\frac{760\times 22.4}{273}=\frac{600V_2}{288}$

$V2=\frac{760\times 22.4\times 288}{273\times 600}$ L

Option D is correct.