Question

The temperature of an ideal gas in a sealed 0.5-m3 rigid container is reduced from 350 K to 270 K. The final pressure of the gas is 70 kPa. The molar heat capacity at constant volume of the gas is 28.0 J/mol · K. The heat absorbed by the gas is closest to_________

Answer 1

Answer : The heat absorbed by the gas is closest to 34.9 kJ

Explanation :

First we have to calculate the moles of gas.

Using ideal gas equation:

`PV=nRT`

where,

P = Pressure of gas = 70 kPa = 70000 Pa

V = Volume of gas =
`0.5m^3`

n = number of moles = ?

R = Gas constant =
`8.314m^3Pa/mol.K`

T = Temperature of gas =
`270K`

Putting values in above equation, we get:

`70000Pa* 0.5m^3=n* (8.314m^3Pa/mol.K)* 270K`

`n=15.59mol`

Heat released at constant volume is known as internal energy.

The formula used for change in internal energy of the gas is:

`\Delta Q=\Delta U\\\\\Delta U=nC_v\Delta T\\\\\Delta Q=nC_v(T_2-T_1)`

where,

`\Delta Q` = heat at constant volume = ?

`\Delta U` = change in internal energy

n = number of moles of gas = 15.59 moles

`C_v` = heat capacity at constant volume gas = 28.0 J/mol.K

`T_1` = initial temperature = 350 K

`T_2` = final temperature = 270 K

Now put all the given values in the above formula, we get:

`\Delta Q=nC_v(T_2-T_1)`

`\Delta Q=(15.59moles)* (28.0J/mol.K)* (270-350)K`

`\Delta Q=-34921.6J=-34.9kJ`

Thus, the heat absorbed by the gas is closest to 34.9 kJ