Question

The pressure of a gas is reduced from 1200.0 mm hg to 850.0 mm hg as the volume of its container is increased by moving a piston from 85.0 ml to 350.0 ml. what would the final temperature be if the original temperature was 90.0 °c?

Answer 1

From gas laws:
$\frac{PV}{T} = Constant$

Therefore,
$\frac{ P_{1} V_{1} }{ T_{1} } = \frac{ P_{2} V_{2} }{ T_{2} }$

P1 = 1200 mm
V1 = 85 ml
T1 = 90°C = 363.15 K
P2 = 850 mm
V2 = 350 ml
T2 = ?

Substituting;
$T_{2} = \frac{ P_{2} V_{2} T_{1} }{ P^{1} V^{1} } = \frac{850*350*363.15}{1200*85} = 1059.19 K$ = 786.04 °C

Answer 2

Using the combined gas law.
P1V1/T1 =P2V2/T2
P1 = 1200 mmHg, V1= 85 ml, T1 = 90 +273 =363 K
P2 = 850 mmHg,   V2 = 350 ml, T2 = ?
T2 = P2V2T1/P1V1
     = ( 850 × 350 ×363)/(1200× 85)
     = 107992500/102000
     = 1058.75 K
The final temperature will be 785.75 °C or 1058.75 K