Question

The temperature of 2.0 mol of a monatomic ideal gas is 340 K. The internal energy of this gas is doubled by the addition of heat. (a) How much heat is needed when it is added at constant volume? J (b) How much heat is needed when it is added at constant pressure?

Answer 1

Answer :

(a) The heat needed when it is added at constant volume is 8.5 kJ.

(b) The heat needed when it is added at constant pressure is 14.1 kJ.

Explanation :

Heat released at constant pressure is known as enthalpy.

Heat released at constant volume is known as internal energy.

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

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

where,

$\Delta Q_v$ = heat at constant volume = ?

$\Delta U$ = change in internal energy

n = number of moles of gas = 2.0 moles

$C_v$ = heat capacity at constant volume for monoatomic gas = $\frac{3}{2}R$

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = 340 K

$T_2$ = final temperature = $340\times 2=680K$

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

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

$\Delta Q_v=(2.0moles)\times (\frac{3}{2}R)\times (680-340)K$

$\Delta Q_v=(2.0moles)\times (\frac{3}{2}\times 8.314J/mole.K)\times (680-340)K$

$\Delta Q_v=8480.28J=8.5kJ$

Thus, the heat needed when it is added at constant volume is 8.5 kJ.

(b) The formula used for change in enthalpy of the gas is:

$\Delta Q_p=\Delta H\\\\\Delta H=nC_p\Delta T\\\\\Delta Q=nC_p(T_2-T_1)$

where,

$\Delta Q_p$ = heat at constant pressure = ?

$\Delta H$ = change in enthalpy energy

n = number of moles of gas = 2.0 moles

$C_p$ = heat capacity at constant pressure for monoatomic gas = $\frac{5}{2}R$

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = 340 K

$T_2$ = final temperature = $340\times 2=680K$

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

$\Delta Q_p=nC_p(T_2-T_1)$

$\Delta Q_p=(2.0moles)\times (\frac{5}{2}R)\times (680-340)K$

$\Delta Q_p=(2.0moles)\times (\frac{5}{2}\times 8.314J/mole.K)\times (680-340)K$

$\Delta Q_p=14133.8J=14.1kJ$

Thus, the heat needed when it is added at constant pressure is 14.1 kJ.