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Snowcat [4.5K]
3 years ago
7

If 500 cal of heat are added to a gas, and the gas expands doing 500 J of work on its surroundings, what is the change in the in

ternal energy of the gas?
Physics
1 answer:
poizon [28]3 years ago
6 0

Answer:

The change in the internal energy of the gas 1,595 J

Explanation:

The first law of thermodynamics establishes that in an isolated system energy is neither created nor destroyed, but undergoes transformations; If mechanical work is applied to a system, its internal energy varies; If the system is not isolated, part of the energy is transformed into heat that can leave or enter the system; and finally an isolated system is an adiabatic system (heat can neither enter nor exit, so no heat transfer takes place.)

This is summarized in the expression:

ΔU= Q - W

where the heat absorbed and the work done by the system on the environment are considered positive.

Taking these considerations into account, in this case:

  • Q= 500 cal= 2,092 J (being 1 cal=4.184 J)
  • W=500 J

Replacing:

ΔU= 2,092 J - 500 J

ΔU= 1,592 J  whose closest answer is 1,595 J

<u><em>The change in the internal energy of the gas 1,595 J </em></u>

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3 years ago
A sample of an ideal gas is in a tank of constant volume. The sample absorbs heat energy so that its temperature changes from 33
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Answer:

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Explanation:

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Before the absorption we have:

(1/2)m\bar{v_{1}}^{2}=\frac{3k_{B}T_{1}}{2} (1)

After the absorption,

(1/2)m\bar{v_{2}}^{2}=\frac{3k_{B}T_{2}}{2} (2)

If we want the ratio of v2/v1, let's divide the equation (2) by the equation (1)

\frac{v_{2}^{2}}{v_{1}^{2}}=\frac{T_{2}}{T_{1}}

\frac{v_{2}}{v_{1}}=\sqrt{\frac{T_{2}}{T_{1}}}

\frac{v_{2}}{v_{1}}=\sqrt{\frac{1340}{335}}

\frac{v_{2}}{v_{1}}=\sqrt{4}

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Answer:

ratio =0.3075 T

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F =  = mv^2/r

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qvB = mv^2/r ...........(i)

B = mv/(rq)  ...............(ii)

If  accelerating potential V is  same and  then  kinetic energy equals the potential energy difference

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 dividing (iv) and (iii) equation we get

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             =0.3075 T

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