Thermodynamics Test - 54

it is given that $$W=80\,J$$

from ideal gas equation

$$ P\Delta V=nR\Delta T $$

$$ \because W=P\Delta T $$

$$ \Delta T=\dfrac{80}{nR}..........(1) $$

Change in internal energy is given by the relation as follows

 $$ \Delta U=n{{c}_{v}}\Delta T $$

$$ =n\dfrac{R}{\gamma -1}\Delta T..........(2) $$

Here, diatomic gas is given so $$\gamma =\dfrac{7}{5}$$

Equation (2) becomes

$$ \Delta U=\dfrac{nR}{\dfrac{2}{5}}\Delta T $$

$$ =\dfrac{5nR}{2}\times \dfrac{80}{nR} $$

$$ =200\,J $$

The heat given to the gas during this process is

$$ \Delta Q=\Delta U+W $$

$$ \Delta Q=200+80 $$

$$ \Delta Q=280\,J $$

The process in which change in volume and temperature of a gas take place at a constant pressure is called an isobaric process.

Consider some gas contained in a barrel. Supply some quantity of heat to increase its temperature from T1 to T2 without changing its pressure ‘P’. It will be observed that the gas expand. The change is said to be isobaric change

$$\textbf {Hint :}$$ Use $$ U=n{{c}_{v}}\Delta T $$

$$\textbf{Step 1: }$$Kinetic energy of molecules is given by $$\dfrac{1}{2}nM{{v}^{2}}$$

Relation between $${{c}_{v}}$$ and internal energy $$U$$is given by

$$ U=n{{c}_{v}}\Delta T $$

$$ {{c}_{v}}=\dfrac{U}{n\Delta T}........(1) $$

$$ \because \,\dfrac{{{c}_{p}}}{{{c}_{v}}}-1=\lambda -1 $$

$$ \dfrac{{{c}_{p}}-{{c}_{v}}}{{{c}_{v}}}=\lambda -1 $$

$$ \dfrac{R}{{{c}_{v}}}=\lambda -1 $$

$$ {{c}_{v}}=\dfrac{R}{\lambda -1}........(2) $$

$$\textbf{Step 2: }$$From (1) and (2)$$ \dfrac{U}{n\Delta T}=\dfrac{R}{\lambda -1} $$

$$ U=\dfrac{Rn\Delta T}{\lambda -1} $$

Now the internal energy is equal to the change in KE

 $$ \dfrac{Rn\Delta T}{\lambda -1}=\dfrac{1}{2}nM{{v}^{2}} $$

$$ \dfrac{R\Delta T}{\lambda -1}=\dfrac{1}{2}nM{{v}^{2}} $$

$$ \Delta T=\dfrac{M{{v}^{2}}}{2R}(\lambda -1) $$If the box suddenly stops, then change in temperature of gas will be $$\Delta T=\dfrac{M{{v}^{2}}}{2R}(\lambda -1)$$

$${\textbf{Correct option: D}}$$