Question

a cell with volume 52 m3 contains gas and a machine. the machine is then turned on remotely, expanding the box to volume of 88 m3 and giving off 830 kj of heat to the gas. the change in internal energy is then determined to be 530 kj. assuming pressure remained constant during this process, what is the pressure of the gas in the cell?

Answer 1

The gases in the cell is under 8333.3 Pa of pressure.

What are different sorts of pressure?

Pressure is the term used to describe the physical force exerted on an object. Per unit area, a perpendicular force is delivered to that same surface of the objects. F/A is the primary formula for tension . Pascals are a unit of pressure (Pa). Relative, atmospheric, disparity, and gauge pressures are various types of pressure.

Briefing:

By using the first law of thermodynamics, we can find the work done by the gas:

ΔU = Q -W

where in this problem:

ΔU = 530kJ  is the change in internal energy of the gas

 Q = 830KJ is the heat absorbed by the gas

W is the work done by the gas (positive if done by the gas, negative otherwise)

Therefore, solving for W,

W = Q - ΔU

   =  830 - 530

   =  300KJ

   =  $3.00 .10^{5}$ J

So, the gas has done positive work: it means it is expanding.

Then we can rewrite the work done by the gas as

$W = p(V_{f} -V_{i} )$

where:

$V_{i} = 52 m^{3}$ is the initial volume of the gas

$V_{f} = 88 m^{3}$ is the final volume

P is the pressure

And solving for P, we find

$V_{f} = V_{i} + \frac{W}{P}$

$88 = 52 + \frac{3.00.10^{5} }{P}$

$88-52 = \frac{3.00.10^{5} }{P}$

$36 = \frac{3.00.10^{5} }{P}$

$P = \frac{3.00.10^{5} }{36}$

$P = 8333.3 Pa$

Therefore , the pressure of the gas is 8333.3 Pa

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