|
3. If `18.75\ mol` of helium gas is at `10.0°C` and a gauge pressure of `0.350\ atm`,
(a) calculate the volume of the helium gas under these conditions.
(b) Calculate the temperature if the gas is compressed to precisely half the volume at a gauge pressure of `1.00\ atm`.
3. If 18.75\ mol of helium gas is at 10.0°C and a gauge pressure of 0.350\ atm,
(a) calculate the volume of the helium gas under these conditions.
(b) Calculate the temperature if the gas is compressed to precisely half the volume at a gauge pressure of 1.00\ atm. `PV=nRT`
a)`V = 0.323\ m^3`
b)
`T_2 = 210\ K or -63°C`
Given:
`n = 18.75\ mol`
`T(K) = 273 +10 = 283\ K`
`P = 1\ atm + P_(gauge) = 1.350\ atm`
known:
`R = 8.315\ J//mol•K`
`1\ atm = 1.013 times 10^5 Pa`
Equation:
`PV=nRT`
(a)
`PV=nRT => V = (nRT)/P`
`V = ((18.75\ mol)(8.315\ J//mol•K)(283\ K))/(1.350\ atm times 1.013 times 10^5 Pa//atm)`
`V = 0.323\ m^3`
(b)
Given: `P_2 = 1.00\ atm+ 1.00\ atm = 2.00\ atm`
Since `n` not changed.
`(P_1V_1)/T_1 = (P_2V_2)/T_2` where `V_2 = 0.5V_1`
`=>`
`(P_1V_1)/T_1 = (P_2 0.5V_1)/T_2`
`P_1/T_1 = (0.5P_2)/T_2`
`T_2 = (0.5P_2)/P_1 T_1`
`T_2 = (0.5 times 2.00\ atm)/(1.35\ atm) 283\ K`
`T_2 = 210\ K or -63°C` |