Suppose 4.00 molof an ideal diatomic gas, with molecular rotation but not oscillation, experienced a temperature increase of 60.0Kunder constant pressure conditions. What are

1. The energy transferred as heat Q .
2. The change $\Delta E_{int}$in internal energy of gas.
3. The work W done by the gas.
4. The change $\Delta K$in total translational kinetic energy of the gas?

No. of moles of a diatomic gas is n=4.00mol

Temperature;

$\mathrm{\Delta T}=60.0\mathrm{K}$

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

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

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 pressure and$\Delta T$is the temperature difference.

The expression for the change in internal energy is given by,

$\Delta E_{int}=n{C}_V\Delta T$

Here $\Delta E_{int}$is the change in internal energy, n is the number of moles, C_Vis the specific heat capacity at constant volume, $\Delta T$is the difference in temperature.

The equation for the energy transferred as heat is given by,

$Q=n{C}_P\Delta T$…… (i)

But, for diatomic gas,

$C_P=\frac{7}{2}R$

Substitute $\frac{7}{2}R$for C_Pinto the equation (i)

$Q=\frac{7}{2}nR\Delta T$

Substitute 4mol for n, 8.3145J/mol.K for R, ${60}^{.}C$for$\Delta T$into the above equation,

$$\begin{gathered} Q=\frac{7}{2}\left(4\right)\left(8.314\right)\left(60\right) \\ =6.98\times {10}^{3}J \end{gathered}$$

Therefore, the amount of energy transferred as heat is.$6.98\times {10}^{3}J$.

The expression for the change in internal energy is given by,

$\Delta E_{int}=n{C}_V\Delta T$…… (ii)

But, for diatomic gas$C_V=\frac{5}{2}R$,

Substitute$\frac{5}{2}R$for $C_V$into the equation (ii)

$\Delta E_{int}=\frac{5}{2}nR\Delta T$

Substitute 4mol for n , 8.3145j/mol.K for R, 60^.C for T into the above equation,

$$\begin{gathered} \Delta E_{int}=\frac{5}{2}\left(4\right)(8.314)\left(60\right) \\ =4.99\times {10}^{3}J \end{gathered}$$

The change in the internal energy of the gas is$4.99\times {10}^{3}J$

According to the first law of thermodynamics,

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

Substitute $6.98x{10}^3]$for Q and $4.99x{10}^3]$for $\Delta E_{int}$into the above equation,

$$\begin{gathered} W=6.98\times {10}^3-4.99\times {10}^3 \\ =1.99\times {10}^{3}J \end{gathered}$$

Therefore the work done by the gas is 1.99$\times$10³J.

The expression for the$\Delta K$in total translational kinetic energy of the gas is given by,

$\Delta K=\frac{3}{2}nR\Delta T$

Substitute 4mol for n, 8.3145J/mol.K for R, 60^.C for$\Delta T$into the above equation,

$$\begin{gathered} \Delta K=\frac{3}{2}\left(4\right)\left(8.314\right)\left(60\right) \\ =2.99\times {10}^{3}J \end{gathered}$$

Therefore, the change in the translational kinetic energy of the gas is 2.99$\times$10³J