Thermodynamics MCQ Questions & Answers in Heat and Thermodynamics | Physics

Work done in expansion is positive and in compression it is negative.

Total internal energy of system = Internal energy of oxygen molecules + Internal energy of argon molecules
$$\eqalign{ & = \frac{{{f_1}}}{2}{n_1}RT + \frac{{{f_2}}}{2}{n_2}RT = \frac{5}{2} \times 2RT + \frac{3}{2} \times 4RT \cr & = 11RT \cr} $$

If we study the $$P - T$$  graph we find $$AB$$  to be a isothermal process, $$AC$$  is adiabatic process given. Also for an expansion process, the slope of adiabatic curve is more (or we can say that the area under the $$P - V$$  graph for isothermal process is more than adiabatic process for same increase in volume).
Only graph (B) fits the above criteria.

A change in pressure and volume of a gas without any change in its temperature is called an isothermal change. In such a change, there is a free exchange of heat between the gas and its surroundings.
$$\therefore T = {\text{constant,}}\,\,\Delta T = 0$$
So, internal energy $$\left( U \right)$$ remains constant is $$\left( {\Delta U = 0} \right).$$

The internal energy of n moles of a gas is
$$U = \frac{1}{2}\,nFRT$$
where $$F$$ = number of degrees of freedom. Internal energy of 2 moles of oxygen at temperature $$T$$ is
$${U_1} = \frac{1}{2} \times 2 \times 5\,RT$$     [$$F$$ = 5 for oxygen molecule]
Internal energy of 4 moles of argon at temperature $$T$$ is
$${U_2} = \frac{1}{2} \times 4 \times 3\,RT = 6\,RT$$
Total internal energy = $$11 \,RT$$

Work is a path function. The remaining three parameters are state function.

In an adiabatic process, $$Q = 0$$
So, from 1^st law of thermodynamics, $$\Delta Q = \Delta U + \Delta W$$
$$\eqalign{ & {\text{As}}\,\,\Delta Q = 0 \cr & {\text{so,}}\,\,W = - \Delta U = - n{C_V}\Delta T \cr & = - n\left( {\frac{R}{{\gamma - 1}}} \right)\left( {{T_f} - {T_i}} \right) \cr & = \frac{{nR}}{{\gamma - 1}}\left( {{T_i} - {T_f}} \right)\,.....\left( {\text{i}} \right) \cr} $$
Given, work done, $$W = 6R\,J,n = 1\;mol,R = 8.31\,J/mol - K,\gamma = \frac{5}{3},{T_i} = TK$$
Substituting given values in Eq. (i), we get
$$\eqalign{ & \therefore 6R = \frac{R}{{\left( {\frac{5}{3} - 1} \right)}}\left( {T - {T_f}} \right) \cr & \Rightarrow 6R = \frac{{3R}}{2}\left( {T - {T_f}} \right) \cr & \Rightarrow T - {T_f} = 4 \cr & \therefore {T_f} = \left( {T - 4} \right)K \cr} $$
NOTE
Adiabatic expansions of mono, dia and polyatomic gases are shown below.
$$1 \to {\text{monoatomic}}\,\,\,2 \to {\text{diatomic}}\,\,\,3 \to {\text{polyatomic}}$$

$$\eqalign{ & Q = W = nRT\ln \frac{{{V_f}}}{{{V_i}}} \cr & T = \frac{Q}{{nR\ln \left( {\frac{{{V_f}}}{{{V_i}}}} \right)}} = \frac{{1500}}{{\frac{{0.5 \times 25}}{{3 \times \ln 3}}}} \cr & \Rightarrow T = \frac{{1500}}{{\frac{{0.5 \times 25}}{{3 \times 1}}}} = 360\,K \cr} $$

$$\eqalign{ & {\text{Process}}\,\left( A \right)\,\frac{{nRT}}{V}{V^2} = K;TV = {\text{constant}};T \propto \frac{1}{V} \cr & {\text{Process}}\,\left( B \right)\,\frac{{nRT}}{V} = K{V^2};\frac{T}{{{V^3}}} = {\text{constant}};T \propto {V^3} \cr} $$

Given, temperature of source,
$$T = {30^ \circ }C = 30 + 273 \Rightarrow {T_1} = 303\,K$$
Temperature of sink, $${T_2} = {4^ \circ }C = 4 + 273$$
$${T_2} = 277\,K$$
As, we know that $$\frac{{{Q_1}}}{{{Q_2}}} = \frac{{{T_1}}}{{{T_2}}} \Rightarrow \frac{{{Q_2} + W}}{{{Q_2}}} = \frac{{{T_1}}}{{{T_2}}}\,\,\left\{ {\because W = {Q_1} - {Q_2}} \right\}$$
where $${{Q_2}}$$ is the amount of heat drawn from the sink (at $${{T_2}}$$ ), $$W$$ is workdone on working substance, $${{Q_1}}$$ is amount of heat rejected to source (at room temperature $${{T_1}}$$ ).
$$\eqalign{ & \Rightarrow W{T_2} + {T_2}{Q_2} = {T_1}{Q_2} \cr & \Rightarrow W{T_2} = {T_1}{Q_2} - {T_2}{Q_2} \cr & \Rightarrow W{T_2} = {Q_2}\left( {{T_1} - {T_2}} \right) \cr & \Rightarrow W = {Q_2}\left( {\frac{{{T_1}}}{{{T_2}}} - 1} \right) \cr & \Rightarrow W = 600 \times 4.2 \times \left( {\frac{{303}}{{277}} - 1} \right) \cr & W = 600 \times 4.2 \times \left( {\frac{{26}}{{277}}} \right) \cr & W = 236.5\,Joules \cr & {\text{Power}} = \frac{{{\text{Work done}}}}{{{\text{Time}}}} = \frac{W}{t} = \frac{{236.5}}{1} = 236.5\,W \cr} $$