Question

The temperature of 2.00molof an ideal monoatomic gas is raised 15.0Kat constant volume. What are

1. The work Wdone by the gas
2. The energy transferred as heat Q
3. The change$\Delta E_{int}$in the internal energy of the gas
4. The change$\Delta K$in average kinetic energy per atom?

Short answer

1. The work done by the gas is zero.
2. The energy transferred as heat Q is 374J.
3. The change $\Delta E_{int}$in the internal energy of the gas is 374J.
4. The change $\Delta K$in the average kinetic energy per atom is 3.11$\times {10}^{-22}J$.

Step-by-step solution

Given

• Number of moles of an ideal monoatomic gas,n=2.00
• Temperature difference,$\Delta T=15.0K$
• Change in volume,$\Delta V=0$

Understanding the concept

The expression for the amount of energy transferred to the gas is given by,

$Q=n{C}_P\Delta T$

Here Q is the amount of energy transferred to gas, n is the number of moles, C_P is the specific heat capacity at constant volume and$\Delta T$is the temperature difference.

The expression for the work done by the gas is given by,

$W=p\Delta V$

Here W is the work done by the gas, p is the pressure and $\Delta V$is the change in the volume.

(a) Calculate the work W done by the gas

Work done by the gas is;

$W=p\Delta V$

But,

$\Delta V=0$

Hence,

W=0

Therefore, the work W done by the gas is zero.

(b) Calculate the energy transferred as heat Q

Heat energy is given by

$Q=n{C}_v\Delta T$$Q=n{C}_v\Delta T$

But, for a monoatomic gas,

$C_v=\frac{3}{2}R$

Substitute role="math" localid="1662468781681" $\frac{3}{2}R$for C_V into the above equation,

$Q=n\left(\frac{3}{2}R\right)\Delta T$
role="math" localid="1662469017230" $$\begin{gathered} =2\left(\frac{3}{2}8.314\right)\left(15.0\right) \\ =374.13 \\ =374J \end{gathered}$$

Therefore, the energy transferred as heat Q is 374J.

(c) Calculate the change in the internal energy of the gas

According to the first law of thermodynamics,

$\Delta E_{int}=Q-W$

Substitute 374J for Q and 0J for W into the above equation,

role="math" localid="1662469480888" $$\begin{gathered} \Delta E_{int}=374-0 \\ =374J \end{gathered}$$

Therefore, the change $\Delta E_{int}$in the internal energy of the gas is 374J.

(d) Calculate the change in average kinetic energy per atom

No. of atoms (N) = No. of moles x Avogadro no.

$\begin{array}{l}N=2\times 6.02\times {10}^{23}\\ =12.04\times {10}^{23}\end{array}$

$Changeintheaveragekineticenergyperatom\left(\Delta K\right)=\frac{\left(\Delta E_{i}nt\right)}{N}$

Substitute 374J for$\Delta E_{i}nt$and $12.04\times {10}^{23}$for into the above equation,

role="math" localid="1662470169132" $\begin{array}{l}K=\frac{374}{12.04\times {10}^{23}}\\ =31.063\times {10}^{-23}\\ =3.11\times {10}^{-22}\end{array}$

Therefore, the change in the average kinetic energy per atom is$3.11\times {10}^{-22}J$