Question #194313

If 4.89 moles of gas are contained in a volume of 47.04 L at 2.61 atm, what is the temperature of the gas in Kelvin?


If 3.02 moles of gas are contained at 2.25 atm and 31 oC, what is the volume of the gas in L?


Calculate the number of moles of gas contained in 11.63 L at 1.64 atm and 19.73 oC.


Calculate the pressure (in atm) on 4.11 moles of a gas contained in 15.19 L at a temperature of 101.54 K.


How many moles of a gas are contained in 48.97 L at a pressure of 2.01 atm and a temperature of 229.32 K?


1
Expert's answer
2021-05-20T07:54:13-0400

(1) Moles, n=4.89n=4.89

Volume, V=47.04 LV=47.04\space L

Pressure, P=2.61 atmP=2.61\space atm

PV=nRTPV=nRT

T=PVnR=2.61×47.044.89×0.0820=306.185 KT=\dfrac{PV}{nR}=\dfrac{2.61\times47.04}{4.89\times0.0820}=306.185\space K


(2) PV=nRTPV=nRT

V=nRTP=3.02×0.0820×(31+273)2.25=22.45 LV=\dfrac{nRT}{P}=\dfrac{3.02\times0.0820\times(31+273)}{2.25}=22.45\space L


(3) PV=nRTPV=nRT

n=PVRT=1.64×11.630.0820×(19.73+273)=0.7945 moln=\dfrac{PV}{RT}=\dfrac{1.64\times11.63}{0.0820\times(19.73+273)}=0.7945\space mol


(4) PV=nRTPV=nRT

P=nRTV=4.11×0.0820×101.5415.19=2.25 atmP=\dfrac{nRT}{V}=\dfrac{4.11\times0.0820\times101.54}{15.19}=2.25\space atm


(5) PV=nRTPV=nRT

n=PVRT=2.01×48.970.0820×229.32=5.23 moln=\dfrac{PV}{RT}=\dfrac{2.01\times48.97}{0.0820\times229.32}=5.23\space mol


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