Question

A gas has a volume of 1.75L at -23C and 150.0kPa. At what temperature would the gas occupy 1.30L at 210.0kPa?

Answer 1

Answer:

T2 = 260 K  

Explanation:

Given data:

P1 = 150.0 k Pa

T1 = (-23+ 273.15) K = 250.15 K  

V1 = 1.75 L  

P2 = 210.0 kPa  

V2 = 1.30 L

To find:

T2 = ?

Formula:

$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$

$T_2 = \frac{P_2 V_2 T_1}{P_1 V_1}$

Calculation:

T2 = (210.0 kPa) x (1.30 L) x (250.15 K) / (150.0 kPa) x (1.75 L)

T2 = 260 K