How many moles of gas are in the bag?

[Wellesley]

Homework Statement

1. A large closed bag of plastic contains 0.10 m3 of an unknown gas at an initial temperature of 10°C and at the same pressure as the surrounding atmosphere, 1.0 atm. You place this bag in the sun and let the gas warm up to 38°C and expand to 0.11 m3. During this process, the gas absorbs 840 cal of heat. Assume the bag is large enough so the gas never strains against it, and therefore remains at a constant pressure of 1.0 atm.

(a) How many moles of gas are in the bag?

(b) What is the work done by the gas in the bag against the atmosphere during the expansion?

(c) What is the change in the internal energy of the gas in the bag?

(d) Is the gas a monoatomic gas? A diatomic gas?

Homework Equations

The Attempt at a Solution

a.)
PV=nRT --> 1 atm * .10 m^3= n * 0.0821 * (10+273)K
n=.1/23.2343--->0.004304 moles

b.)
W=P*deltaV--->1 atm*.01m^3 = .01 Joule

c.) delta U= Delta Q - delta W --> (840 cal*4.186J)-0.01 J
delta U=3516.24 J-0.01J=3516.23J

d.)
delta Q = nc_p*delta T --> 3516.23J=0.004304 mol * c_p*28K
c_p=29177.4 Joules


I think part D is where my problem lies. The bold number I think is especially off. Am I approaching this problem the right way? My book doesn't give any good examples...Thanks.

 

[rock.freak667]

b.)
W=P*deltaV--->1 atm*.01m^3 = .01 Joule

You mean W=PΔV = 1atm * 0.01m³= 101.325(10³)Pa * 0.01m³, this will give you more than 0.01J of energy. Your method is correct, it just in this part, you forgot to convert units.

 

[Wellesley]

You mean W=PΔV = 1atm * 0.01m³= 101.325(10³)Pa * 0.01m³, this will give you more than 0.01J of energy. Your method is correct, it just in this part, you forgot to convert units.

Thanks. I knew it was something simple.
But my answer is still off (I think) by a factor of 1000!

ΔQ = ncp*ΔT --> 2502.99=0.004304 mol * c_p*28K
c_p=20769.6 Joules/mol*K --> 4961.69 cal/mol*k

When I looked up the c_p values, they were ~ 4.96cal/mol*K. But I've checked my math over several times. Am I missing something basic?

 

[rock.freak667]

The Attempt at a Solution

a.)
PV=nRT --> 1 atm * .10 m^3= n * 0.0821 * (10+273)K
n=.1/23.2343--->0.004304 moles

Sorry, I missed it here as well, P=101.325 kPa, redo this and get 'n'.

Thanks. I knew it was something simple.
But my answer is still off (I think) by a factor of 1000!

ΔQ = ncp*ΔT --> 2502.99=0.004304 mol * c_p*28K
c_p=20769.6 Joules/mol*K --> 4961.69 cal/mol*k

When I looked up the c_p values, they were ~ 4.96cal/mol*K. But I've checked my math over several times. Am I missing something basic?

 

[Wellesley]

Sorry, I missed it here as well, P=101.325 kPa, redo this and get 'n'.

I always get confused with 8.31 (typically physics) and 0.0821 (chemistry) as the gas constants.

It turns out neither of them use m³ ...that turns into 100L and I get 4.3 moles, and 4.96 cal/mol*K

Thanks for the help!