Answer to Question #166457 in General Chemistry for Angela

Question

A 5.00-L air sample has a pressure of 107 kPa at a temperature of –50.0ºC. If the temperature is raised to 102ºC and the pressure drops to 60.0 kPa, what will the new volume be?

Answer 1

There are two stages of gas:

At first stage

$V_1=$ 5 L

$P_1$$=$ 107kPa

$T_1=$ $-50$ degre cecsius= $-$ 50$+$ 273$=$ 223 K

At second stage:

$P_2$ =60kPa

$T_2=$ 102 degre celsius = 102$+$ 273$=$ 375 K

$V_2=V_2$ (let)

From gas eqn $\Rightarrow$ PV $=$ nRT

$\Rightarrow$ n $=\dfrac{PV}{RT}$

So here in question moles are constant at both stages,so:

$\Rightarrow$ $n_1=n_2$

$\Rightarrow$ $\dfrac{P_1V_1}{RT_1}=\dfrac{P_2V_2}{RT_2}$

R is conatant from both the sides so

$\Rightarrow$ $V_2=\dfrac{P_1V_1T_2}{T_1P_2}$

$\Rightarrow$ $V_2$ $=\dfrac{107\times5\times375}{223\times60}$ $=$ 14.99 L$=$ 15.00 L approximtely