Calculating Work Done by Ideal Gas at Constant Pressure: PV Diagram Analysis

[Punkyc7]

Two moles of an ideal gas are heated at constant pressure from a temperature of 29C to 110C.
calculate the work done by the gas


So work is the area under the PV curve, the only problem is I have no volume or intital pressure so PV=nRT doesn't help. Then I was thinking about Q= delta U -W but that doesn't seem to help.


How should I go about this problem because I am stuck?

 

[Pi-Bond]

The work done in general is

$W=\int PdV$

With constant pressure, this simplifies to

$W=P(V_{f}-V{i})$

where V_f and V_i are the final and initial volumes respectively. You can find these volumes by using the fact that the pressure P is constant (both volumes will have P in the denominator which will be eliminated from the expression above)

 

[rock.freak667]

Two moles of an ideal gas are heated at constant pressure from a temperature of 29C to 110C.
calculate the work done by the gas


So work is the area under the PV curve, the only problem is I have no volume or intital pressure so PV=nRT doesn't help. Then I was thinking about Q= delta U -W but that doesn't seem to help.


How should I go about this problem because I am stuck?

Well if the work done by a gas at constant pressure is given by W=P(V₂-V₁) how would this translate if you replace PV with nRT? (with the appropriate subscript)

 

[zspdztzx]

Two moles of an ideal gas are heated at constant pressure from a temperature of 29C to 110C.
calculate the work done by the gasSo work is the area under the PV curve, the only problem is I have no volume or intital pressure so PV=nRT doesn't help. Then I was thinking about Q= delta U -W but that doesn't seem to help.How should I go about this problem because I am stuck?

Hi! Here are my two cents!

If there is no heat loss or particle loss in this process, then given the pressure is constant, therefore internal energy of the gas, delta-U, is zero, which, translates to, heat that flowed in, delta-Q, equals to W, the work done by the gas (i.e. expansion).

Therefore, avg KE increase in gas (reflected as temp increase) shall equal to the heat energy that flowed in, in turns, shall equal to the work done by the gas in expansion.

Equationally speaking:
W=Q=Delta-KE=n*(3/2)*R*Delta-T
(where "n" is number of moles of gas, "R" is ideal gas constant, and "Delta-T" is temperature difference in K and it's equal to temp difference in C).Maybe I am wrong (please correct me if so!).. :)