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

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

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

Put the value into the formula

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

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

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

Using formula of heat

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

Put the value into the formula

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

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

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.