Question

A 0.220 g sample of carbon dioxide has a volume of 0.50L and pressure of 0.60 atm. What is the temperature of the gas in K? (R = 0.0821)

Answer 1

T=731.0 K

Step-by-step explanation:

According to the Ideal Gas Law, pV=nRT where

p=pressure in atm

V=volume in L

n=moles of gas

R=Ideal Gas Constant (0.08205
`(L*atm)/(mol*K)`)

T=temperature in K

To calculate the moles of carbon dioxide, first calculate the molecular weight:

`C=12.01(g)/(mol)\\O=16.00(g)/(mol)`

So,

`MW\ CO_2=12.01(g)/(mol)+2(16.00(g)/(mol))=44.01(g)/(mol)`

Then use the molecular weight to convert grams to moles:

`0.220g\ CO_2((1\ mol\ CO_2)/(44.01\ g\ CO_2))=5.0x10^(-3)\ mol\ CO_2`

So, for this problem, let

p = 0.60 atm

V = 0.50 L

n =
`5.0x10^(-3)` moles

R = 0.0821
`(L*atm)/(mol*K)`

So,

`(0.60\ atm)(0.50\ L)=(5.0x10^(-3)mol)(0.0821(L*atm)/(mol*K))T\\T=((0.60\ atm)(0.50\ L))/((5.0x10^(-3)mol)(0.0821(L*atm)/(mol*K)))\\T=731.0\ K`