Question

Method 5 of W Heat taken from a gas in a process is 80 J and increase in internal energy of the gas is 20J. Find work done by the gas in the given process.

Answer 1

$\huge\underline{\underline{\bf \orange{Question}}}$

Heat taken from a gas in a process is 80 J and increase in internal energy of the gas is 20J. Find work done by the gas in the given process.

$\huge\underline{\underline{\bf \orange{Solution-}}}$

$\large\underline{\underline{\sf Given:}}$

• Heat taken (Q) = 180 J
• Internal Energy (U) = 20 J

$\large\underline{\underline{\sf To\:Find:}}$

• Work done (W) .

❏$\large{\boxed{\bf \blue{\triangle Q= \triangle U + \triangle W}}}$

$\implies{\sf 180=20+W }$

$\implies{\sf W = 180-120 }$

$\implies{\bf \red{Work\:Done(W) = 160 J }}$

$\huge\underline{\underline{\bf \orange{Answer-}}}$

Work done by gas ${\bf \red{160J}}$

Answer 2

Work done by the gas = -100 J

Given:

Heat taken from a gas in a process is 80 J and increase in internal energy of the gas is 20J.

To find:

Work done by the gas in the given process.

Concept used:

First law of thermodynamics

$\Delta Q = \Delta U + \Delta W$

$\Delta Q$ = Heat absorbed by gas

$\Delta U$ = Change in internal energy

$\Delta W$ = Work done by the gas

Explanation:

Heat taken from a gas in a process,

So that $\Delta Q$ = -80j

Increase in internal energy of the gas is 20J.

So that $\Delta U$ = 20J

From first law of thermodynamics,

$\Delta Q = \Delta U + \Delta W$

-80 = 20 + $\Delta W$

$\Delta W$ = -100J

Negative sign indicate that work done on the gas.

So Work done by the gas = -100 J

To learn more...

1) Name the physical quantity that remains conserved in first law of thermodynamics.

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2)Why second law of thermodynamics is known as law of degradation of energy.

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