Question

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?

Answer 1

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

Step-by-step 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

`\Delta H_(1)=nC_(p)*\Delta T`

Put the value into the formula

`\Delta H_(1)=3*6.9*(320-300)`

`\Delta H_(1)=414\ cal`

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

Using formula of heat

`\Delta H_(1)=nC_(v)*\Delta T`

Put the value into the formula

`\Delta H_(1)=3*4.9*(300-320)`

`\Delta H_(1)=-294\ cal`

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

Using formula of total heat

`\Delta H_(T)=\Delta H_(1)+\Delta H_(2)`

`\Delta H_(T)=414-294`

`\Delta H_(T)=120\ cal`

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