Question #c4606

1 Answer

"24 L"24 L

Explanation:

NIST uses a temperature of 20^@"C"20C ("293.15 K"(293.15 K, 68^@"F")68F) and an absolute pressure of "1 atm"1 atm ("14.696 psi"(14.696 psi, "101.325 kPa")101.325 kPa). This standard is also called normal temperature and pressure (abbreviated as NTP).

Normal Temperature and Pressure

"T = 293.15K"T = 293.15K
P = 1 atmP=1atm

so

The only difference between STP and NTP is that NTP conditions are at 20^@"C" = "293.15 K"20C=293.15 K, rather than at 0^@"C" = "273.15 K"0C=273.15 K.

Since only the temperature is different, use Charle's Law:

V_1/T_1 = V_2/T_2V1T1=V2T2

We have:

V_1 = 22.41 L V1=22.41L

T_1 = 273.15 K T1=273.15K

V_2 = ? larr "This is the volume of 1 mol of a gas at NTP "V2=?This is the volume of 1 mol of a gas at NTP

T_2 = 293.15 K. T2=293.15K.

Plugging our information into Charle's Law:

V_1/T_1 = V_2/T_2V1T1=V2T2

=> (22.41 L)/(273.15 K) = V_2/(293.15 K)22.41L273.15K=V2293.15K

=> V_2 = (22.41 L)(293.15 K)/(273.15 K) = 24.05 LV2=(22.41L)293.15K273.15K=24.05L

"Therefore, 1 mol of any gas will occupy about 24 L at NTP."Therefore, 1 mol of any gas will occupy about 24 L at NTP.

Hope this helped