You can use the Ideal Gas Law or the Combined Gas Law. PV/T = PV/T ---- left are usually labeled (1) and right (2) to show different conditions left P = unknown, V = 19.6L, T = (273+27)K right P= 760 torr or 1 amt, V = 22.4 L, T = 273K --- all are standard P(19.6L)/300K = 1atm(22.4L)/273K --- solve for P P = (1 atm)(22.4L)(300K)/ [(273K)(19.6L)] do the math --- get the answer.
This is equivalent to 0.2 atm. (152 mmHg)*(1 atm)/(760 mmHg)=0.2atm
0.045 mol L 0.5. Mol.
Standard temperature and pressure (STP) is a standard by which comparisons can be made. STP is 0°C (273 K) and 1.00 ATM (760 mmHg, 101.325 kPa). Molar volume is based upon the conditions at STP, which is 22.4 L for 1 mole of any [ideal] gas.
PV=nRT. So at ntp P=1bar. If n= 1mole. R= .082. T=273K. Then 1V=1.273 .082. v=22.44 l of gas
Answer: 597.4 ATM PV=nRT P=8(0.08206)(273K)/0.3L P=597.4 ATM
You can use the Ideal Gas Law or the Combined Gas Law. PV/T = PV/T ---- left are usually labeled (1) and right (2) to show different conditions left P = unknown, V = 19.6L, T = (273+27)K right P= 760 torr or 1 amt, V = 22.4 L, T = 273K --- all are standard P(19.6L)/300K = 1atm(22.4L)/273K --- solve for P P = (1 atm)(22.4L)(300K)/ [(273K)(19.6L)] do the math --- get the answer.
Decreasing the pressure of a gas will increase its volume -- C
Krypton is a gas at STP 1 atm. and 25oC
-273K
water changes from a gas to a solid to a liquid
solid
what is the molecular weigh of a gas if 15L of its vapor at 100C and 1 atm weigh 68g
The total pressure will be 5 atm and the partial pressure of gas 1 will be 2 atm and the partial pressure of gas 2 will be 3 atm.
Standard temperature and pressure conditions are: Temp: 273K or 0°C Pressure: 100k or 1 bar STP is 273K or 0oC and 1 bar or 100kPa
1.0
The lowest possible temperature is -273.15 C which equated to '0'K. Therefore, temperature -273K cannot exist. No -ve K temperatures exist.