(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Pt. A) A gas is compressed at a constant pressure of 0.632 atm from 7.11 L to 5 L. In the process,

440 J of energy leaves the gas by heat.

What is the work done on the gas?

Answer in units of J

Pt. B)

What is the change in its internal energy?

Answer in units of J

2. Relevant equations

PV = nRT

U = Q + W

1 atm = 101325 pa

1L = .001m^3

3. The attempt at a solution

A)

.632atm = 64037.4 pa

7.22L = .00711 m^3

5L = .005 m^3

W = P(Vf -Vi)

=

64037.4(.005-.00711)

=

-135.189

U = Q+W

U = -135.189 -440

=

-575.1189

OR (done other ways)

: 304.881

: -304.881

: 575.118914

All of these were wrong. Please help?

B) ...

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# Homework Help: Work done on compressed gas

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