- #1
theaterlord
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Hello all, i am haveing a few issues with a thermodynamics/thermal problem and was wondering if someone could lend a fresh set of eyes to it.
a gas is contained in a cylinder with a frictionless piston the helium is compressed quasistatically from an initial state where the volume is 4.23*10^-3 m^3 and the pressure is Pi*10^5 Nm^-2 to a final state in which the volume is Vf*10^-3 m^3 . the Temp remains constant at 280K during the process, and the gas approximates to an ideal gas.
a) find the final pressure of the system
b) find the work done on the gas
c) the change in the internal energy of the gas
d) the heat flow into the system durring the process
e) the entropy change of the gas
A&B seem to be fine i have issues with the assumptions that need to be made for c/d/e
A)
as the pressure is low boyle's law applys Quick Symbols
B)
finding the work done
C)
Q+W=0
W=-Q
D)
E)
using the entropy form of the first law
for a quasistatic reversible isothermal process
du=0
Du=dQ+dW
but
dQ=T.ds
dw=-P.dv
so
dU=Tds-Pdv=0
p=constant/v
ds=c/(t*v) dv
to which we integrate
\Delta S=c/t \int^{vf}_{vi} dv/v
which gives
So i think A,B and E are ok, however i am not certain about Questions c&d
If someone could cast their eyes over this and tell me what they think i would be very grateful!
Homework Statement
a gas is contained in a cylinder with a frictionless piston the helium is compressed quasistatically from an initial state where the volume is 4.23*10^-3 m^3 and the pressure is Pi*10^5 Nm^-2 to a final state in which the volume is Vf*10^-3 m^3 . the Temp remains constant at 280K during the process, and the gas approximates to an ideal gas.
a) find the final pressure of the system
b) find the work done on the gas
c) the change in the internal energy of the gas
d) the heat flow into the system durring the process
e) the entropy change of the gas
Homework Equations
The Attempt at a Solution
A&B seem to be fine i have issues with the assumptions that need to be made for c/d/e
- compressed quasistaticly
- temp is constant thus isothermal
- the system is isolated as it does not interact with the surroundings
- it behaves as an ideal gas
A)
as the pressure is low boyle's law applys Quick Symbols
Vi Pi=Vf Pf
thus Pf=(PiVi)/Vf
B)
finding the work done
W=- \int^{vf}_{vi} Pdv
P V= constant
thusP V= constant
P=Constant/v
W=- constant \int^{vf}_{vi} dv/v
w=-constant ln(vi/Vf)
W=- constant \int^{vf}_{vi} dv/v
w=-constant ln(vi/Vf)
C)
\Delta U= Q+W
\Delta U = 0
process is closed
\Delta U = 0\Delta U = 0
process is closed
Q+W=0
W=-Q
D)
\Delta U= Q+W
process is closed
\Delta U = 0
Q+W=0
W=-Q
process is closed
\Delta U = 0
Q+W=0
W=-Q
E)
using the entropy form of the first law
for a quasistatic reversible isothermal process
du=0
Du=dQ+dW
but
dQ=T.ds
dw=-P.dv
so
dU=Tds-Pdv=0
p=constant/v
ds=c/(t*v) dv
to which we integrate
\Delta S=c/t \int^{vf}_{vi} dv/v
which gives
\Delta S=(c/t) ln(vf/Vc)
If someone could cast their eyes over this and tell me what they think i would be very grateful!
