How do i do this Chemistry problem?

  • Thread starter Thread starter xxran96xx
  • Start date Start date
X

xxran96xx

Guest
This is probably the most useless subject ever but i need help with this question so can someone help me
here it is

The Gas in a closed container has a pressure of 3.00 X 10^2 kPa at 30 degrees Celsius (303 K). What will the pressure be if the temperature is lowered to -172 Degrees Celcius (101 K)?

Can someone explain what all these symbols mean and explain when you solve please and if you could help with this problem too

Calculate the Volume of a gas (in L) at a pressure of 1.00 X 10^2 kPa if it's volume at 1.20 X 10^2 is 1.50 X 10^3mL

Again Please explain all these symbols
 
You have to use the formula: P1xV1/T1=P2xV2/T2
P: Pressure
V: Volume
T: Temperature

(note: if you are not given one of these, assume it is constant, in which case it will cancel out, so you can omit that variable from the formula)
 
Use Gay Lussac's Law, P1T2=P2T1

1. Given:
P1=3.00 X 10^2 kPa
T1=303 K
T2=101 K

P1T2=P2T1
3.00 X 10^2 kPa (101 K) = P2 (303 K)
P2= 1.00 X10^2 kPa

Use Boyle's LAw P1V1=P2V2
2. Given:
P1=1.00 X 10^2 kPa
P2=1.20 X 10^2 kPa
V2=1.50 X 10^3mL

P1V1=P2V2
1.00 X 10^2 kPa (V1)= 1.20 X 10^2 kPa (1.50 X 10^3mL)
V1= 1.80 X10^3 mL
 
Back
Top