What is the work done when a sample of gas expands from 12.5L to 17.2L against a pressure of 1.29 atm? Give your answer in Joules.

Answer 1

The work done is 614.33 Joules.

When a gas expands at constant pressure then for a small change in volume #'dv'# workdone is #dw=pdv#. If the volume changes from #'v_1'# to #'v_2'# at constant pressure #'p'#, the work done is #dw=p(v_2-v_1)#. When work is done by the system against external pressure then #dw=pdv# #rArrw=int_(v_1)^(v_2) p dv# So, by calculating with given numericals we get, #rArr dw=1.29atm(17.2L-12.5L)=1.29atmxx(4.7L)=6.063atm.L#.
To have the answer in terms of we use below units: #1 atm = 101325 Pa#
But#1 Pa = 1 J//m^3# #(#because #J=Pa.m^3##)#.
Substitute '#1Pa#' value to get #'1atm#' value, #rArr 1atm=101325J//m^3#. Now as we know #1L#itre=#10^-3m^3#.
Converting the workdone #'dw'# value into Joules using the above required units we get, #dw=6.063xx101325J/cancel(m^3)xx10^-3cancel(m^3)=614.3334J#. #:.# work done in terms of Joules is #614.33J#oules.
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Answer 2

The work done is 470.43 joules.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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