Question

A 208.1 mL sample of gas exerts 574.6 mm Hg pressure at 44.4 ºC. What pressure does it exert at 68.1 ºC if the volume expands to 401.5 mL?
a. 320
b. 457
c. 190
d. 917

Answer 1

Answer:

a.  P = 320 mmHg

Explanation:

For this problem, the pressure, volume, and temperature are changing, so we'll need to combine Boyle's Law and Charles' Law:

$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}$

For this equation, the temperatures must be measured in Kelvin.  The rest of units in the equation only need to match between beginning and end conditions.

Recall that to convert from Celsius to Kelvin, add 273, or use the equation $T_C+273=T_K$.

So $T_1=(44.4+273)[K]=317.4[K]$  and $T_2=(68.1+273)[K]=341.1[K]$

Substituting known values, we can solve for the unknown:

$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}$

$\dfrac{(574.6[mmHg])(208.1[mL])}{(317.4[K])}=\dfrac{P_2(401.5[mL])}{(341.1[K])}$

$\dfrac{(574.6[mmHg])(208.1[mL]\!\!\!\!\!\!\!\!\!\!{--} )}{317.4[K]\!\!\!\!\!\!{-}}*\dfrac{341.1[K] \!\!\!\!\!\!{-}}{401.5[mL] \!\!\!\!\!\!\!\!\!\!\!{--} }=\dfrac{P_2(401.5[mL] \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{-----})}{341.1[K] \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----}}*\dfrac{341.1[K] \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----}}{401.5[mL] \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{-----} }$

$320.056719296[mmHg]=P_2$

Accounting for significant figures, $P_2=320.1[mmHg]$.  

The closest answer provided is 320, so "a".