absolute pressure=101.3kpa(273-20)/273*1500/1000
Gauge pressure=absolute-atmospheric.
determine
So just estiamting, answer b is in the ballpark.
1500 cm3 of ideal gas at STP is cooled to -20°C and put into a 1000 cm3 container. What is the final gauge pressure?
11 kPa
40 kPa
113 kPa
141 kPa
240 kPa
2 answers
a liter is 10^3 = 1000 cm^3
so we have 22.4 * 1.5 = 33.6 moles of gas at 273 deg K and 100 kPa
P V = n R T
n and r remain the same
so
P1V1/T1 = P2V2/T2
100 (1500)/273 = P2 (1000)/253
P2 = 139 kPa ABSOLUTE pressure
- 100 KPa for gage
= 39 k Pa gage
so we have 22.4 * 1.5 = 33.6 moles of gas at 273 deg K and 100 kPa
P V = n R T
n and r remain the same
so
P1V1/T1 = P2V2/T2
100 (1500)/273 = P2 (1000)/253
P2 = 139 kPa ABSOLUTE pressure
- 100 KPa for gage
= 39 k Pa gage
1 answer