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Physics_S16
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Please Help! Thermodynamics, the ideal gas equation and temperature change
Hi all! I'm new here! I hope my question gets answered because I'm really confused.
An amount of 0.18 mol of an ideal gas is held in an insulated cylinder fitted with a piston.
Atmospheric pressure is 1.0 x 10^5 Pa
The volume of the gas is suddenly increased from 1.8 x 10^3 cm3 to 2.1 x 10^3 cm3.
For the expansion of the gas,
(i) calculate the work done by the gas and hence show that the internal energy changes by 30J
(ii)determine the temperature change of the gas and state whether the change is an increase or a decrease
Δu = q + w
pV=nRT?
(i) work done by gas= pressure xΔV
So, work done by gas = 30J
Therefore, Δu = -30J
Part (ii) is the problem here
(ii) 3/2 x nRΔT = Δu
So, ΔT = (30x2)/(3x0.18x8.31) = 13.4 K (decrease) ?
However, when using pV=nRT and re-arranging it such that ΔT=(pΔV)/(nR), the
answer is different and it is an increase
Thanks in advance! Any help is much appreciated!
Hi all! I'm new here! I hope my question gets answered because I'm really confused.
Homework Statement
An amount of 0.18 mol of an ideal gas is held in an insulated cylinder fitted with a piston.
Atmospheric pressure is 1.0 x 10^5 Pa
The volume of the gas is suddenly increased from 1.8 x 10^3 cm3 to 2.1 x 10^3 cm3.
For the expansion of the gas,
(i) calculate the work done by the gas and hence show that the internal energy changes by 30J
(ii)determine the temperature change of the gas and state whether the change is an increase or a decrease
Homework Equations
Δu = q + w
pV=nRT?
The Attempt at a Solution
(i) work done by gas= pressure xΔV
So, work done by gas = 30J
Therefore, Δu = -30J
Part (ii) is the problem here
(ii) 3/2 x nRΔT = Δu
So, ΔT = (30x2)/(3x0.18x8.31) = 13.4 K (decrease) ?
However, when using pV=nRT and re-arranging it such that ΔT=(pΔV)/(nR), the
answer is different and it is an increase
Thanks in advance! Any help is much appreciated!