Question 1

A mixture of 0.5 mol N _2 gas, 1.0 mol O _2 gas and 1.5 mol H _2 gas exerts a total pressure of 18 bar. Find the partial pressure of O _ 2(g) .

Accepted Answer

Total number of moles in the mixture is: n_ total = n_ N_2 + n_ O_2 + n_ H_2 n_ total = 0.5 + 1.0 + 1.5 = 3.0 mol Mole fraction of O _2 is: x_ O_2 = n_ O_2 n_ total = 1.0 3.0 = 1 3 According to Dalton's law of partial pressures, the partial pressure of O _2 is: P_ O_2 = x_ O_2 P_ total P_ O_2 = 1 3 18 = 6.0 bar Answer: 6.0 bar

Question 2

Find the number of moles of CO _2 present in a sample occuping 4 10^ -3 \, m ^3 at 1.104 10^5\, Nm ^ -2 pressure (R T = 2208\ J mol ^ -1 )

Accepted Answer

Using the ideal gas equation, PV = nRT Rearranging for the number of moles n, we get n = PV RT Substituting the given values: n = 1.104 10^5 4 10^ -3 2208 n = 441.6 2208 n = 0.2 mole Answer: 0.2 mole

Question 3

Which of the following is not an illustration of viscosity?

Accepted Answer

Gradation of lubricant oils is done on the basis of their viscosity. High viscosity of blood is an indicator of cardiovascular diseases. Glass is a supercooled liquid with very high viscosity, which causes it to flow downwards very slowly over time, resulting in the thickening of glass panes at the bottom in old buildings. The cleansing action of soap is based on the principle of surface tension, not viscosity. Soap lowers the surface tension of water, allowing it to spread and penetrate more ea

Question 4

A container contains mixture of He and Ne gases at certain temperature. He is 20 % by mass of mixture. What is the ratio of partial pressure of Helium to Neon ?

Accepted Answer

Let the total mass of the mixture be 100 g. Mass of He = 20 g Mass of Ne = 80 g Molar mass of He = 4 g/mol Molar mass of Ne = 20 g/mol Number of moles of He, n_ He = 20 4 = 5 Number of moles of Ne, n_ Ne = 80 20 = 4 According to Dalton's law of partial pressures, the ratio of partial pressures is equal to the ratio of number of moles. P_ He P_ Ne = n_ He n_ Ne = 5 4 Answer: 5:4

Question 5

What ratio by mass of 'Ne' and CH _4 should be mixed so that partial pressure exerted by each gas is same?

Accepted Answer

Let the mass of Ne be w_1 and the mass of CH _4 be w_2. Molar mass of Ne = 20 g/mol Molar mass of CH _4 = 16 g/mol Number of moles of Ne, n_ Ne = w_1 20 Number of moles of CH _4, n_ CH _4 = w_2 16 According to Dalton's law of partial pressures, the partial pressure of a gas in a mixture is directly proportional to its number of moles. Since the partial pressures of both gases are equal, their number of moles must be equal. n_ Ne = n_ CH _4 w_1 20 = w_2 16 w_1 w_2 = 20 16 = 5 4 The required ratio

Question 6

A gas at 10\,^ C occupies a volume of 283 mL. if it is heated to 20\,^ C keeping the pressure constant. What is the new volume of gas ?

Accepted Answer

According to Charles's Law, at constant pressure, V_1 T_1 = V_2 T_2 Given: V_1 = 283 mL T_1 = 10\,^ C = 10 + 273 = 283 K T_2 = 20\,^ C = 20 + 273 = 293 K Substituting the values: 283 283 = V_2 293 V_2 = 293 mL Answer: 293 mL

Question 7

V_0 and V_t are volumes of an ideal gas at 0\ ^ C and t\ ^ C respectively, find ratio V_t V_0

Accepted Answer

According to Charles's law, at constant pressure, the volume of a fixed mass of an ideal gas is directly proportional to its absolute temperature. V T V_t V_0 = T_t T_0 The absolute temperatures are T_0 = 273 K and T_t = (273 + t) K . V_t V_0 = 273 + t 273 V_t V_0 = 1 + t 273 Answer: 1 + t 273

Question 8

At constant temperature, a gas occupies a volume of 200 mL at a pressure of 500 mm Hg. What will be its volume at 800 mm Hg pressure?

Accepted Answer

According to Boyle's law, at constant temperature, P_1 V_1 = P_2 V_2. Given P_1 = 500\ mm Hg , V_1 = 200\ mL , and P_2 = 800\ mm Hg . Substituting the values: 500 200 = 800 V_2 V_2 = 500 200 800 = 125\ mL Answer: 125\ mL

Question 9

Which of the following statements is true about ideal gas ?

Accepted Answer

An ideal gas is a hypothetical concept and does not exist in reality. Real gases behave as ideal gases only under conditions of low pressure and high temperature. According to the kinetic theory of gases, an ideal gas strictly obeys Boyle's law, Charles' law, and other gas laws under all conditions of temperature and pressure. The collisions between its molecules are perfectly elastic, and there are no intermolecular forces of attraction between its molecules. Thus, the statement that it does no

Question 10

Calculate the new pressure of the gas enclosed in a cylinder when it compressed from 5 L to 2 L at 1 atm and 300 K. The final temperature in this process is 500 K.

Accepted Answer

Using the combined gas law equation: P_1 V_1 T_1 = P_2 V_2 T_2 Given values are: P_1 = 1 atm, V_1 = 5 L, T_1 = 300 K V_2 = 2 L, T_2 = 500 K Substituting the values into the equation: 1 5 300 = P_2 2 500 P_2 = 1 5 500 300 2 P_2 = 25 6 4.16 atm Answer: 4.16 atm

Question 11

Calculate the compressibility factor of 1 mole of a certain real gas if it occupies 0.4 dm ^3 at 300 K and 40 atm. [R = 0.082 atm dm ^3 K ^ -1 mol ^ -1 ]

Accepted Answer

The compressibility factor Z is given by the formula: Z = PV nRT Substituting the given values: Z = 40 0.4 1 0.082 300 Z = 16 24.6 Z 0.65 Answer: 0.65

Question 12

A gas occupies a volume of 0.35 dm^3 at 290 K and 2 atm. Pressure. Calculate the volume of gas at STP.

Accepted Answer

Using the combined gas equation: P_1 V_1 T_1 = P_2 V_2 T_2 Given: P_1 = 2 atm V_1 = 0.35 dm^3 T_1 = 290 K At STP: P_2 = 1 atm T_2 = 273 K Substituting the values: 2 0.35 290 = 1 V_2 273 V_2 = 2 0.35 273 290 V_2 = 191.1 290 0.66 dm^3 Answer: 0.66 dm^3

Question 13

Calculate the temperature of 2 moles of gas occupying a volume of 5.0 liters at 24.6 atmospheres? (R = 0.0821 L atm K^ -1 mol^ -1 )

Accepted Answer

Using the ideal gas equation PV = nRT Substituting the given values: 24.6 5.0 = 2 0.0821 T 123 = 0.1642 T T = 123 0.1642 = 749.08 K Converting to Celsius: T = 749.08 - 273.15 = 475.93^ C 476^ C Answer: 476^

Question 14

A syringe has a volume of 10.0\ cm ^3 at a pressure of 1 atm. If the end is sealed and the plunger is pushed down (constant temperature), what will be the final volume when the pressure becomes 3.5 at

Accepted Answer

Using Boyle's Law at constant temperature: P_1 V_1 = P_2 V_2 Given P_1 = 1\ atm , V_1 = 10.0\ cm ^3, and P_2 = 3.5\ atm . Substituting the values: 1 10.0 = 3.5 V_2 V_2 = 10.0 3.5 = 2.857\ cm ^3 Rounding to appropriate significant figures, V_2 = 2.86\ cm ^3. Answer: 2.86\ cm ^3

Question 15

According to the kinetic molecular theory of gases, the average kinetic energy of gas molecules:

Accepted Answer

According to the kinetic molecular theory of gases, the average translational kinetic energy of a gas molecule is given by E = 3 2 kT, where k is the Boltzmann constant and T is the absolute temperature. For one mole of an ideal gas, the average kinetic energy is E = 3 2 RT, where R is the universal gas constant. This shows that the average kinetic energy of gas molecules depends only on the absolute temperature and is directly proportional to it. It is independent of pressure, volume, and the n

States of Matter — Previous Year Questions & Solutions