Helium gas with a volume of 2.85 L, under a pressure of 1.28 atm and at a temper

Question

Helium gas with a volume of 2.85 L, under a pressure of 1.28 atm and at a temperature of 43.5 C, is warmed until both pressure and volume are doubled.

1.) What is the final temperature?

Express your answer using three significant figures.

Tf = ?? K

2.) How many grams of helium are there? The molar mass of helium is 4.00g/mol.

Express your answer using three significant figures.

m = ?? g

I have no idea how to do these problems. Please show step bt step with correct answers so I can try to understand.

THANK YOU!!

Explanation / Answer

its simple u only need to use the idal gas equation

PV = nRT (n = no. of moles of gas)

as no. of moles do not change we can write n = PV/RT = constant

so we use P1V1/T1 = P2V2/T2

we know P1 =1.28 atm V1 = 2.85 litres T1 = 43.5 C = 43.5 + 273.15 = 316.65 K

P2 = 2*P1 = 2.56 and V2 = 2*V1 = 5.7 litres

puting in equation 1.28*2.85/316.65 = 2.56*5.7/T2

T2 = 1266.6 K =1.26*103 ----------------------- ( using 3 significant figures)

now to find mass we need no. of moles

n = P1V1/RT1

= 1.28*2.85/8.20*10-2316.65 -----------------(value of R is 8.20*10-2 L atm K-1mol-1)

n = 0.1404 moles

now wt in grams = n*molec mass = 0.1404*4

= 0.5619 grams

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