## Class 11 Chemistry First Law Of Thermodynamics

**Question 1. Calculate die change in internal energy when heat released by the die system Is 200 J and work done on the die system is 120J heat absorbed and work done by the system arc 200J and 120J respectively 0heat released and work done by the system arc 200 J and 120 J respectively.**

**Answer:** As per given data, q = -200 J and w = +120 J

∴ Change in internal energy,

AU = q + w =(- 200 + 120) J =-80 J

As per given data, q = +200 J and w = -120

∴ Change in internal energy,

AU = q+w=(200-120) J = + 80 J

As per given data, q = -200 J and w = -120 J

∴ Change in internal energy,

AU = q + w=(- 200- 120) J = -320 J

**Question 2. In a process, the system performs 142 J of work and the internal energy of the system Increases by 879 J. Predict the direction of heat flow and also calculate the quantity of heat transferred.**

**Answer:** Given, w = -142 J and AU = + 879 J

∴ AU = q+w or, 879 = q- 142

∴ q = + 1021 J

The positive value of q indicates that the system absorbs heat from the surroundings i.e., heat flows from the surroundings to the system. The amount of heat transferred in the process is 1 021 J.

**Question 3. A system undergoes n process in which It gives up 300J of heat and Its internal energy decreases by 300J. Of the system and its surroundings, which one docs work in this process?**

**Answer:** Given, A U = -300J and q = -900 J

∴ AU = q + w or, -300 = – 900 + w or, w = + 600 J

w is (+)ve. This means work is done on the system by the surroundings and the amount of work done is 600 J

**Question 4. The initial volume of a gas, confined in a cylinder fitted with a piston is U.2L. The final volume of the gas becomes 33.6L after expansion against a constant external pressure of 2atm. During expansion if the gas absorbs 1kg of heat from the surroundings then what will be the change in the internal energy of that gas?**

**Answer:** Work done, w = -Pex(V2-V t)

Given: Pgx = 2 atm; Vj = 11.2 L and V2 = 33.6 L

∴ w =-2(33.6- 11.2) = -44.8 L-atm = -44.8 X 101.3

= -4538.24 J [v 1 L-atm = 101.3 J]

So, the amount of work done by the gas is 4538.24 J.

Again, the heat (q) absorbed by the gas = 1 of = 1000J Therefore, the change in internal energy of the gas, AU = q+w =(1000-4538.24) J = -3538.24 J

Therefore, the decrease in internal energy is 3538.24 J

**Question 5. An ideal gas is expanded from 1L to 6L in a closed vessel at 2 atm pressure by applying heat at a fixed temperature. Calculate the work done and heat absorbed by the gas. [Given: lLatm = 24.22 cal]**

**Answer:** Work done, w = -PexAV = -Pex(V2- V1)

Given: Pgx = 2 atm; Vy = 1 L and V2 = 6 L

w = -2(6-1) = -10 L-atm =-242.2 cal

From the first law of thermodynamics, AU = q + w.

Since the expansion is carried out isothermally and the system is an ideal gas, the change in internal energy in the process will be equal to zero. Thus, AU = 0.

0 = q + w or, q = -w = + 242.2cal

Therefore, the amount of heat absorbed bythe gas is 242.2 cal.

**Question 6. A cylinder fitted with a piston contains an ideal gas with a volume of 21L. The gas is compressed isothermally to l/3rd of its initial volume under a constant external pressure of 3 atm. Calculate q, w, and AU.**

**Answer:** Work done, w = -Pex(V2- Vx) [Vy > V2]

Given: Pgx = 3 atm; Vy = 21L and V2 ,\(=21 \times \frac{1}{3}=7 \mathrm{~L}\)

∴ w = -3(7-21) = 42 L. atm

= 42 X 101.3 J = 4254.6 J p[since 1L.atm =101.3J]

In an isothermal process of an ideal gas, the internal energy ofthe gas does not change i.e., AU = 0

From the first law of thermodynamics, AU = q + w

∴ 0 = q + 4254.6 J or, q = -4254.6 J

Thus, in the process g=- 4254.6 J, w=+ 4254.6 J, AU=0.