A 1-m-long monopole car radio antenna operates in the AM frequency of 1.5 MHz. How muchcurrent is required to transmit 4 W of po

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ValentinkaMS [3465]

Answer:

The amount of heat that enters the gas throughout this two-step process totals 120 cal.

Explanation:

Given that,

Moles present = 3

Heat capacity at volume held constant = 4.9 cal/mol.K

Heat capacity at pressure held constant = 6.9 cal/mol.K

Starting temperature = 300 K

Ending temperature = 320 K

We are tasked with determining the heat absorbed by the gas at constant pressure

Employing the heat formula

$\Delta H_{1}=nC_{p}\times\Delta T$

Substituting the values into the equation

$\Delta H_{1}=3\times6.9\times(320-300)$

$\Delta H_{1}=414\ cal$

Next, we calculate the heat absorbed by the gas at constant volume

Using the corresponding heat formula

$\Delta H_{1}=nC_{v}\times\Delta T$

Insert the values into the formula

$\Delta H_{1}=3\times4.9\times(300-320)$

$\Delta H_{1}=-294\ cal$

Now, it's necessary to evaluate the total heat flow into the gas during both steps

Using the total heat formula

$\Delta H_{T}=\Delta H_{1}+\Delta H_{2}$

$\Delta H_{T}=414-294$

$\Delta H_{T}=120\ cal$

Thus, the heat that transfers into the gas throughout this two-step process amounts to 120 cal.

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