OUNT OF HEAT THAT FLOWS DURING THE EXPANSION

[coffeem]

Homework Statement
3 Moles of a monotomatic ideal gas undergo an isothermal (T=50 Degrees Clecius) reversible expansion from a volume of 3m^3 and a pressure P1 to a volume V2 and a pressure P2. The gas does 9.7kJ of work in the expansion. Find:

a) the pressure P1
b) The volume V2
c) The pressure P2
d) The amound of heat that flows during the expansion

The attempt at a solution

My attempt is as follows:

n = 3
T = 323K
V1 = 3m^3

a) P1V1 = nRT

rearanging:

P1 = 2680Pa

b) I am slightly more stuck on:

for an isothermal expansion: dW = -PdV
by subbing into the ideal gas eq" of state:

W = nRTln(V1/V2)

Now since change in T = 0, therefore the chance in E = 0.

and first law of thermodynamics: E = Q +W
therefore Q = -W

so Can i just rearrange with this to find V2 from the expression for work? thanks

 

[ehild]

Yes, but take care, the work done by the gas is given, and you wrote the formula for the external work, done on the gas.

ehild

 

[Andrew Mason]

for an isothermal expansion: dW = -PdV

In this case, dW would be the work done ON the gas. The convention is to use W to represent work done BY the gas, in which case dW = PdV. Note: this is true for any process, not just isothermal.

by subbing into the ideal gas eq" of state:

W = nRTln(V1/V2)

This is true only if the process is isothermal (constant T). Again, you are using W as the work done on the gas which is not the conventional way to express W. W = nRTln(V2/V1) gives youi the work done BY the gas.

Now since change in T = 0, therefore the change in E = 0.

and first law of thermodynamics: E = Q +W
therefore Q = -W

Again, you should use the convention for W as the work done BY the gas. Q = E (or U) + W or E = Q - W where W is the work done BY the gas and Q is the heat flow INTO the gas.

so Can i just rearrange with this to find V2 from the expression for work? thanks

Yes. But keep the signs straight. (Use $Q = W = \int PdV = nRT\ln{(V_2/V_1)}$).

AM