Q: A container has two chambers of volumes V_1=2 litres and V_2=3 litres separated by a partition made of a thermal insulator. The chambers contains n_1=5 and n_2=4 moles of ideal gas at pressures p_1=1

By energy conservation aligned & E _1+ E _2= E _ mix \\ & n _1 f _1 RT _1 2 + n _2 f _2 RT _2 2 = ( n _1 n _2 ) fRT _ mix 2 \\ & 3 2 P _1 ~V _1+ 3 2 P _2 ~V _2= 3 2 P _ mix ( V _1+ V _2 ) \\ & 3 2 1 2+ 3 2 2 3= 3 2 P _ mix (5) \\ & P _ mix = 8 5 \\ & P _ mix =1.6 ~atm aligned

Question 1

One mole of an ideal monatomic gas undergoes a cyclic process as shown in the figure. The total heat supplied to the gas is :

Accepted Answer

For a complete cyclic process, the change in internal energy is zero, i.e., U = 0. According to the first law of thermodynamics, Q = U + W. Therefore, the net heat supplied to the gas is equal to the net work done by the gas: Q = W. The net work done W is given by the area enclosed by the cycle on the P-V diagram. W = Area of rectangle abcd W = (P_a - P_d) (V_b - V_a) W = (300 - 100) (5 - 2) W = 200 3 = 600 J Thus, the total heat supplied to the gas is 600 J . Answer: 600 J

Question 2

An electric heater supplies heat to a system at a rate of 100 W. If the system performs work at a rate of 75 J/s, then the rate at which internal energy increases will be :

Accepted Answer

According to the first law of thermodynamics: dQ dt = dU dt + dW dt Given that the rate of heat supplied is dQ dt = 100 W and the rate of work done by the system is dW dt = 75 J/s = 75 W. Substituting the values: 100 = dU dt + 75 dU dt = 100 - 75 = 25 W The rate at which internal energy increases is 25 W. Answer: 25 W

Question 3

In an adiabatic expansion, the temperature of one mole of an ideal monatomic gas ( =5/3) decreases from 60K to 50K. The work done by the gas in the process is : (Take the universal gas constant as R=8

Accepted Answer

The work done by an ideal gas in an adiabatic process is given by the formula: W = nR(T_1 - T_2) - 1 Given values are: n = 1 mole R = 8.3 J mol^ -1 K^ -1 T_1 = 60 K T_2 = 50 K = 5 3 Substituting these values into the formula: W = 1 8.3 (60 - 50) 5 3 - 1 W = 8.3 10 2 3 W = 83 3 2 W = 249 2 = 124.5 J Answer: 124.5 J

Question 4

A container has two chambers of volumes V_1=2 litres and V_2=3 litres separated by a partition made of a thermal insulator. The chambers contains n_1=5 and n_2=4 moles of ideal gas at pressures p_1=1

Accepted Answer

By energy conservation aligned & E _1+ E _2= E _ mix \ & n _1 f _1 RT _1 2 + n _2 f _2 RT _2 2 = ( n _1 n _2 ) fRT _ mix 2 \ & 3 2 P _1 ~V _1+ 3 2 P _2 ~V _2= 3 2 P _ mix ( V _1+ V _2 ) \ & 3 2 1 2+ 3 2 2 3= 3 2 P _ mix (5) \ & P _ mix = 8 5 \ & P _ mix =1.6 ~atm aligned

Question 5

Two gases A and B are filled at the same pressure in separate cylinders with movable pistons of radius r_A and r_B, respectively. On supplying an equal amount of heat to both the systems reversibly un

Accepted Answer

According to question Pressure is constant P_A=P_B aligned & Q _ A = Q _ B \ & U _ A = U _ B aligned So according to I law aligned & Q = U + w \ & w _1= W _2 \ & P _1 ~V _1= P _2 ~V _2 \ & ( r _ A ^2 ) x _ B = ( r _ B ^2 ) x _ A \ & r _ A r _ B = x _ B x _ A = 9 16 = 3 4 aligned

Question 6

The equilibrium state of a thermodynamic system is described by A. Pressure B. Total heat C. Temperature D. Volume E. Work done

Accepted Answer

Equilibrium state of a thermodynamic system is defined by state variables P, V and T. Work done and total heat are path variables.

Question 7

A thermodynamic system is taken through the cycle a b c d a. The work done by the gas along the path b c is:

Accepted Answer

Path b c is an isochoric process. Work done by gas along path b c is zero.

Thermodynamics — Previous Year Questions & Solutions