Question

A gas is confined to a vertical cylinder by a piston of mass 2 kg and radius 1 cm. When 5J of heat are added, the piston rises by 2.4 cm. Find: (a) the work done by the gas; (b) the change in its internal energy. Atmospheric pressure is 105Pa.​

Answer 1

Answer:

a) Work done by the gas is 1.23 J

b) Change in internal energy of the gas is 3.77 J

Explanation:

Part a)

As per FBD of the piston we can say that

$P_oA + mg = P_gA$

so here we will have

$P_o + \frac{mg}{A} = P_g$

now work done by the gas is given as

$W = P_g\Delta V$

$W = (P_o + \frac{mg}{A})\Delta V$

$W = (P_oA + mg)x$

$W = (10^5(\pi(0.01)^2) + 2\times 10)(0.024)$

$W = 1.23 J$

Part b)

Now we know by first law of thermodynamics

$Q = \Delta U + W$

so we have

$5 = \Delta U + 1.23$

$5 - 1.23 = \Delta U$

$\Delta U = 3.77 J$

#Learn

Topic : Thermodynamics

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