51.
In a Carnot engine, the temperature of reservoir is $${927^ \circ }C$$ and that of sink is $${27^ \circ }C.$$ If the work done by the engine when it transfers heat from reservoir to sink is $$12.6 \times {10^6}J,$$ the quantity of heat absorbed by the engine from the reservoir is
52.
A Carnot engine having an efficiency of $$\frac{1}{{10}}$$ as heat engine, is used as a refrigerator. If the work done on the system is $$10\,J,$$ the amount of energy absorbed from the reservoir at lower temperature is
Consider schematic diagram for a Carnot engine as shown below.
In case of engine,
$${\text{engine efficiency}} = \frac{{{\text{work}}}}{{{\text{heat absorbed}}}} = \frac{W}{{{q_1}}}$$
$$\eqalign{
& \therefore \frac{W}{{{q_1}}} = \frac{1}{{10}} \Rightarrow \frac{{10\,J}}{{{q_1}}} = \frac{1}{{10}} \cr
& {\text{or}}\,\,{q_1} = 100\,J \cr} $$
When this engine is reversed, it takes in work $$W$$ and heat $${q_2}$$ from cold reservoir and ejects $$100\,J$$ of heat to hot reservoir.
$$\eqalign{
& \therefore W + {q_2} = {q_1} \cr
& \Rightarrow 10 + {q_2} = 100 \cr
& {\text{or}}\,\,{q_2} = 90\,J \cr} $$
53.
Which of the following processes is irreversible?
Concept
Apply first law of thermodynamics. According to first law of thermodynamics
$$\Delta U = \Delta Q + \Delta W$$
For isothermal process, $$\Delta U = 0$$
So, $$\Delta Q = - \Delta W$$
Given, $$\Delta W = - 150\,J$$
So, $$\Delta Q = + 150\,J$$
When $$Q$$ is positive, the heat is added to the gas.
55.
Unit mass of a liquid with volume $${V_1}$$ is completely changed into a gas of volume $${V_2}$$ at a constant external pressure $$P$$ and temperature $$T.$$ If the latent heat of evaporation for the given mass is $$L,$$ then the increase in the internal energy of the system is
$$\eqalign{
& Q = mL = 1 \times L = L;W = P\left( {{V_2} - {V_1}} \right) \cr
& {\text{Now }}Q = \Delta U + W \cr
& {\text{or}}\,\,L = \Delta U + P\left( {{V_2} - {V_1}} \right) \cr
& \therefore \Delta U = L - P\left( {{V_2} - {V_1}} \right) \cr} $$
56.
A gas is expanded from volume $${V_0}$$ to $$2{V_0}$$ under three different processes. Process 1 is isobaric process, process 2 is isothermal and process 3 is adiabatic.
Let $$\Delta {U_1},\Delta {U_2}$$ and $$\Delta {U_3}$$ be the change in internal energy of the gas is these three processes. Then -
This is a statement of second law of thermodynamics
58.
For an ideal gas four processes are marked as 1,2,3 and 4 on $$P-V$$ diagram as shown in figure. The amount of heat supplied to the gas in the process 1, 2, 3 and 4 are $${Q_1},{Q_2},{Q_3}$$ and $${Q_4}$$ respectively, then correct order of heat supplied to the gas is - [$$AB$$ is process-1, $$AC$$ is process-2, $$AD$$ is adiabatic process-3 and $$AE$$ is process-4]
In process-1 heat supplied $$=$$ area under $$AB$$ curve + $$n \times {c_v} \times 100\,\left( {{\text{isobaric process}}} \right)$$
In process-2 heat supplied $$=$$ area under $$AC$$ curve (isothermal process)
In process-3 heat supplied $$= 0$$ (adiabatic process)
In process-4 heat supplied $$ = n \times {c_v}\,\left( {T - 600} \right)\,\left( {{\text{isobaric process}}} \right)$$
59.
An ideal gas undergoes a quasi static, reversible process in which its molar heat capacity $$C$$ remains constant. If during this process the relation of pressure $$P$$ and volume $$V$$ is given by $$P{V^n}$$ = constant, then $$n$$ is given by (Here $${C_P}$$ and $${C_V}$$ are molar specific heat at constant pressure and constant volume, respectively):
60.
A monoatomic gas at a pressure $$p,$$ having a volume $$V$$ expands isothermally to a volume $$2 V$$ and then adiabatically to a volume $$16 V.$$ The final pressure of the gas is (take $$\gamma = \frac{5}{3}$$ )