1) 0.03 m^3 of a gas are initially at a pressure of 250 kPa. The pressure of the

Question

1) 0.03 m^3 of a gas are initially at a pressure of 250 kPa. The pressure of the gas is increased at constant temperature to 800kPa. What is the final volume of the gas?

2) A gas is enclosed in a container fitted with a piston of cross-=sectional area 0.150 m2. The pressure of the gas is maintained at 5000 Pa as the piston moves inward 20.0 cm. a) Calculate the work done by the gas. b) If the internal energy of the gas decreases by 8.00 J, find the amount of heat removed from the system during the compression.

Explanation / Answer

PART 1:-
Initial volume,V(i)= 0.03 m^3

Final volume,V(f)= to calculate

Initial Pressure,p(i)= 250kPa

Final Pressure,p(f)= 800kPa

n,R & T will be constant.
So, putting these values in formula pV=nRT,
we get,

250*(.03)=nRT----1

& 800*V(f)=nRT----2
Dividing 2 by 1, we get,

V(f)= 9.375*10^(-3) m^3= 0.0093 m^3 which is the final volume of the gas.

PART 2:-

In an isobaric process, such as this one, work is given by
W = P*?V

To find ?V, use the formula for the volume of a cylinder:
V = A*L

For a change in L:
?V = A*?L

Thus:
W = P*A*?L---- 1

P=5000 Pa; A:=0.150 m^2; ?L:= -0.2 m (negative indicates decrease)

So, putting the values in equation 1, we get, Work done by gas, W= -150 Joules (Here negative indicates work is done by the surroundings on the gas )

(B) The first law of thermodynamics, with an accepted sign convention for heat and work:

Q = ?U + W

Substitute:
Q = ?U + P*A*?L--- 2

?U := -8 J

So, putting the values in equation 2, we get,
amount of heat removed from the system during the compression, Q= -158 Joules (Here negative indicates heat is transfered out of the system)

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