91Ó°ÊÓ

Nitrogen gas $=0.10mol$

For the isotherm process:

The work done the energy transferred between the environment and the system. When it is negative, it means the energy is transferred from the gas to the environment and positive when the energy is transferred from the environment to the gas.

This is an isothermal ideal-gas process where the temperature is constant. So, the work done is the area under the $pV$curve either geometrically or by carrying out the integration which is given by equation (19.13) in the form,

$Q=W=-nRT\ln \left(\frac{V_2}{V_1}\right)$

We are not given the value of the temperature, so from the ideal-gas law, we can get an expression for $T$by

$p_1{V}_1=nRT$

$T=\frac{p_1{V}_1}{nR}$

Now, we use the expression of $p$into equation (1) to get the heat by

$Q=-nR\left(\frac{p_1{V}_1}{nR}\right)\ln \left(\frac{V_2}{V_1}\right)=-p_1{V}_1\ln \left(\frac{V_2}{V_1}\right)$

localid="1648620542421" $=-\left(1.013×{10}^5\mathrm{Pa}\right)\left(3000×{10}^{-6}{\mathrm{m}}^3\right)\ln \left(\frac{1000×{10}^{-6}{\mathrm{m}}^3}{3000×{10}^{-6}{\mathrm{m}}^3}\right)$

$=333\mathrm{J}$

For the adiabatic process:

The adiabatic process is the process where the heat is absorbed or released by the system is zero $Q=0$. But it doesn't mean that the temperature changes.

So, no heat is absorbed from the surrounding. So, the heat is zero

$Q_{\text{adiabtic}}=0\mathrm{J}$

Hence, the heat required for $Q_{\text{Isotherm}}=333\mathrm{J}$and$Q_{\mathrm{adiabtic}}=0\mathrm{J}.$