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:

$70000Pa\times 0.5m^3=n\times (8.314m^3Pa/mol.K)\times 270K$

$n=15.59mol$

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:

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

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

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

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