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!).. :)

 

[rwisz]

To calculate the work done by an ideal gas at constant pressure, we can use the formula W = PΔV, where P is the constant pressure and ΔV is the change in volume. In this case, since the pressure is constant, we can also use the formula W = PΔV = nRTΔV, where n is the number of moles, R is the gas constant, and T is the change in temperature.

In this scenario, we are given the temperature change from 29C to 110C, but we still need to determine the change in volume. This can be found by using the ideal gas law, PV = nRT, and solving for V. We can then plug in the values for n, R, and the initial and final temperatures, to find the change in volume.

Once we have the change in volume, we can plug it into the formula W = PΔV to calculate the work done by the gas. It is important to note that the work done by the gas in this scenario is equal to the amount of heat transferred to the gas, as the pressure is constant and no other forms of work are being done.

In summary, to calculate the work done by an ideal gas at constant pressure, we can use the formula W = PΔV or W = nRTΔV, where we determine the change in volume using the ideal gas law. This allows us to find the amount of work done by the gas during the given temperature change.