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Three moles of an ideal gas with a molar heat capacity at constant volume of 4.9 cal/(mol∙K) and a molar heat capacity at constant pressure of 6.9 cal/(mol∙K) starts at 300 K and is heated at constant pressure to 320 K, then cooled at constant volume to its original temperature. How much heat flows into the gas during this two-step process?

Respuesta :

Answer:

The heat flows into the gas during this two-step process is 120 cal.

Explanation:

Given that,

Number of moles = 3

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

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

Initial temperature = 300 K

Final temperature = 320 K

We need to calculate the heat flow in to gas at constant pressure

Using formula of heat

[tex]\Delta H_{1}=nC_{p}\times\Delta T[/tex]

Put the value into the formula

[tex]\Delta H_{1}=3\times6.9\times(320-300)[/tex]

[tex]\Delta H_{1}=414\ cal[/tex]

We need to calculate the heat flow in to gas at constant volume

Using formula of heat

[tex]\Delta H_{1}=nC_{v}\times\Delta T[/tex]

Put the value into the formula

[tex]\Delta H_{1}=3\times4.9\times(300-320)[/tex]

[tex]\Delta H_{1}=-294\ cal[/tex]

We need to calculate the heat flows into the gas during two steps

Using formula of total heat

[tex]\Delta H_{T}=\Delta H_{1}+\Delta H_{2}[/tex]

[tex]\Delta H_{T}=414-294[/tex]

[tex]\Delta H_{T}=120\ cal[/tex]

Hence, The heat flows into the gas during this two-step process is 120 cal.

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