Q:7. A cylinder with a movable piston contains 3 moles of H_{2} at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston in insulated by having a pile of sand on it. By what factor does the pressure of the gas increase, if the gas is compressed to half of its original volume ? (Given γ = 1.4)

Sol:

The process is adiabatic, as no heat in exchanged

$\displaystyle P_2 V_2 ^\gamma = P_1 V_1 ^\gamma $

$ \displaystyle \frac{P_2}{P_1} = (\frac{V_1}{V_2})^\gamma $

Given V_{2} = V_{1}/2

$\displaystyle \frac{P_2}{P_1} = (\frac{V_1}{V_1/2})^{1.4} $

= 2.64

Q: 8. In changing state of a gas adiabatically from an equilibrium state A to another equilibrium state B, an amount of work equal to 22.3J is done on the system. If the gas is taken from state A to B via a pross in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in the latter case (take 1 cal = 4.19 J)?

Sol:

For change in adiabatic, ΔQ = 0, ΔW = -22.3 J

Now ΔU = ΔQ -ΔW = 0 – (-22.3) = 22.3 J

Second time, ΔQ = 9.35 cal = 9.35 × 4.2 = 39.3 J

ΔW = ΔQ -ΔU = 39.3 – 22.3 = 17 J

Q: 9.An electric heater supplies heat to a system at a rate of 100 W. If system performs work at a rate of 75 joules per second, at what rate is the internal energy increasing?

Sol:

Given, ΔQ = 100W = 100 J/s

Useful work done, ΔW = 75 Js-1

Now, ΔU = ΔQ -ΔW = 100 – 75 = 25 J/s.

Q: 10. Two cylinders A and B of equal capacity are connected to each other via stop crock. The cylinder a contains an ideal gas at standard temperature and pressure, while the cylinder B is completely evacuated. The entire system is thermally insulated. The stop crock is suddenly opened. Answer the following:

(a) What is he final pressure of the gas A and B ?

(b) What is the change in internal energy of the gas?

(c) What is the change in temperature of the gas?

(d) Do the intermediate state of the system (before setting to the final equilibrium state) lie on its P-V-T surface ?

Sol:

(a) If the stop crock opens suddenly, volume of gas would become double at 1 atmospheric pressure, therefore, pressure would become half (i.e. 1/2 atmosphere).

(b) As no work is done on the system by the gas, there will be no change in internal energy.

(c) Gas does not do any work on expanding, thereby, no change in temperature of gas would occur.

(d) No, the process of the expansion is rapid and cannot be controlled. The intermediate state would be a non-equilibrium state, which do not follow gas equation. In due course, gas will come back to equilibrium state.

Q:11. A refrigerator is to remove heat from the eatable kept inside it at 10°C, Calculate the coefficient of performance, if room temperature is 36°C.

Sol:

Given T_{1} = 36°C = 36 + 273 = 309 K;

T_{2} = 10°C = 10 + 273 = 283°K

Coefficient of performance $ \displaystyle = \frac{T_2}{T_1 – T_2} $

$ \displaystyle = \frac{283}{309 – 283} $

=10.9

Q:12. A steam engine delivers 5.4 × 10^{8} J of work per minute and absorbs 3.6 × 10^{9} J of heat per minute from its boiler. What is the efficiency of the engine? How much heat is wasted power minute?

Sol:

Useful work done per min (Out put) = 5.4 × 10^{8} J ;

Heat absorbed per min (Input) = 3.6 × 10^{9} J

Efficiency = Output/Input

=(5.4×10^{8})/(3.6×10^{9} )

=0.15 = 15%

Heat energy wasted/min = Heat absorbed/ min – useful work done/ minute

= 3.6 × 10^{9} – 5.4 × 10^{8}

= 3.06 × 10^{9} J