NEET Practice Questions (MCQs) with Answers & Solutions

Heat given to a system ($∆Q$) is equal to the sum of increase in the internal energy $\left(∆U\right)$ and the work done $\left(∆W\right)$ by the system against the surrounding and 1 cal=4.2J.

According to first law of thermodynamics

            $∆U=Q-W$

                 = $2\times 4.2\times 1000-500$

                = $8400-500$

                = 7900 J

For an adiabatic process there should not be any exchange of heat between the system and its surroundings. All walls of the container must be perfectly insulated. In adiabatic changes gases obey Poisson's law, i.e, $p{V}^{\gamma }=cons\tan t.$ In an isochoric process volume remains constant and for isobaric process pressure remains constant.

In a cyclic process, system is not isolated from the surroundings.

In a cyclic process, a system starts from one point and ends at the same point. In this case, the change in the internal energy must again be zero and therefore the thermal energy added to the system must equal the work done during the cycle. That is, in a cyclic process.

                  $∆U=0 or E=0$

Writing ideal gas law 

                $$\begin{gathered} PV=nRT \\ or PV=\frac{m}{M}RT \\ or \frac{PV}{m}=\frac{1}{M}RT \\ or \frac{P}{\rho }=\frac{RT}{M} \\ or \frac{\rho }{P}=\frac{1}{T} \\ \therefore \frac{{\rho }_1/P_2}{{\rho }_2/P_1}=\frac{T_2}{T_1} \\ \Rightarrow \frac{x}{\left({\rho }_1/P_2\right)}=\frac{383}{283} \\ or \frac{{\rho }_2}{P_2}=\frac{283}{383}x \end{gathered}$$

(d) 

 $$\begin{gathered} From the ideal gas equation\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2} \\ \frac{E_2}{E_1}=\frac{P_2}{P_1}=\frac{V_1}{V_2}\times \frac{T_2}{T_1}=\left(\frac{1}{4}\right)\times \left(\frac{400}{300}\right)=\frac{1}{3}\Rightarrow E_2=\frac{E_1}{3} \\ i.e. elasticity will become \frac{1}{3}times. \end{gathered}$$

Change in internal energy (ΔU) depends upon initial an find state of the function while ΔQ and ΔW are path dependent also.

This is the case of free expansion and in this case $\Delta W=0$, $\Delta U=0$ so temperature remains same i.e. 300 K.

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

⇒ $\Delta W=\Delta Q-\Delta U=100-40=70J$

Work done is not a thermodynamical function.

$\Delta Q=\Delta U+\Delta W=167+333=500cal$