it jee 2006 sample question paper

8/3/2019 IT JEE 2006 Sample Question Paper

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IT JEE 2006 Sample Question PaperPassage 1

When heat is added to a gas, the temperature increases and the gas expands. If the gas isin a rigid container, then the volume remains constant and the pressure increases. If thegas in in a container that can expand (for example a cylinder with a vertical piston), thenthe pressure remains constant but the volume expands.

The amount of energy required to increase the temperature of one mole of gas is calledthe specific heat; there is a specific heat for constant pressure (Cp) and a specific heat for constant volume ((Cv).

In equation form, this is written:

Delta Q is the energy required to raise the temperature of the gas by a certain amount(delta T). The specific heats for a few gases, including monoatomic gases such as heliumand argon, and diatomic gases such as nitrogen and oxygen, are given in the table below.

Specific Heat of Some GasesGas Cv Cp Cp-CvHelium(H) 3.00 4.98 1.98Argon(A) 3.00 5.00 2.00

Nitrogen(N 2) 4.96 6.95 1.99Oxygen(O 2) 4.96 6.95 1.99Carbon Monoxide(CO) 4.93 6.95 2.02Carbon Dioxide(CO 2) 6.74 8.75 2.01Methane(CH 2) 6.48 8.49 2.01

Note: All quanties are in calories(Kelvin*mole)

Q. 1 Let X be the energy needed to raise the temperature of 5 moles of nitrogen held atconstant pressure by one degree. Let Y be the energy needed to raise 5 moles of carbonmonoxide by one degree with the pressure held constant. What is the ratio X:Y?

A. 5:7B. 1:1C. 7:5D. 7:9


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Answer: B

Discription: There is no need to do any calculations here. If you put X on top of Y, youwill see that it reduces to the ratio of Cp for nitrogen over Cp for carbon monoxide. Andsince both gases have the same Cp, the ratio is simply 1:1.

Q. 2 The difference Cp - Cv is a constant. This constant is often called R, the universalgas constant. Which of the following is true given the data?

A. For a monoatomic gas, Cp = 3/2 R B. For a diatomic gas, Cp = 3/2 R C. For a monoatomic gas, Cv = 3/2 R D. For a diatomic gas, Cv = 3/2 R

Answer: C

Discription: Helium and argon are monoatomic gases, and we can see that Cv = 3/2 *R =(3/2)*2 = 3.

Q. 3 How much energy would be required to heat two moles of methane by one degree if the gas is kept at constant volume?

A. 6.5 caloriesB. 8.5 caloriesC. 11 caloriesD. 13 calories

Answer: D

Description: The definition of Cv is the amount of energy required to heat one mole of agas by one degree. Therefore, to heat two moles of methane by one degree will require2*6.5 = 13 calories.

Q. 4 Which of the following is a possible explanation for the fact that Cp is alwaysgreater than Cv?

A. Some of the energy is used to expand the container in order to maintain constant pressure.

B. A rigid container does not conduct heat as well as one that can change shape.C. There are generally more moles of gas when the pressure is kept constant thanwhen the volume is kept constant.

D. There are generally fewer moles of gas when the pressure is kept constant thanwhen the volume is kep constant.

Answer: A


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Discription: Because Cp is always greater than Cv we know that it takes more energy toincrease a given amount of gas when the pressure is held constant. It is reasonable thatthe extra energy is used to increase the volume of the container.

Q. 5 A certain amount of energy, X, is sufficient to raise the temperature of 60 moles of

argon by T degrees when the pressure is constant. How many moles of argon can beraised by T degrees with the same amount of energy X, if the volume is held constant?

A. 30B. 50C. 75D. 100

Answer: D

Discussion: You can use the two equations:

You are given that X = 60*5*T. Let n be the number of moles that can be heated totemperature T with X calories of energy. Then X = n*3*T, and for this equation to beconsistent with the previous one, it must be that n = 100.

Passage 2

When food is cooked, only some of the nutrients are preserved. Food scientists performedtwo experiments to study the retention of vitamin C after potatoes were cooked. Thescientists measured the retention of ascorbic acid (AA) and dehydroascorbic acid (DAA).The sum of the amounts of AA and DAA equals the total amount of Vitamin C.

Experiment 1 In this experiment, potatoes were peeled and then fried. The following table shows thestability of ascorbic acid and dehydroascorbic acid after frying. The concentrations of AAand DAA are given in milligrams per 100 grams of dry potato.

SampleDAA

(mg/100g)AA

(mg/100g)

Total Content of vitamin C(mg/100g)

Raw, peeled potatoes before frying 7.4 44.6 52.0

Fried potatoes140 o for 10 min140 o for 20 min140 o for 30 min180 o for 5 min

29.733.742.742.8

20.67.300

50.341.042.742.8


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Experiment 2 In this experiment, potatoes were peeled and then boiled in water with varyingconcentrations of NaCl. The following table shows the stability of ascorbic acid (AA)anddehydroascorbic acid (DAA) after boiling.

SampleDAA

(mg/100g)AA

(mg/100g)Total Content of

vitamin C(mg/100g)

Raw peeled potatoesCoocked in water Coocked in 1%

NaClCoocked in 5%

NaClCoocked in 10%

NaCl

7.49.09.17.15.8

43.117.113.111.28.9

50.526.122.218.314.7

Q. 6 Which of the following statements accurately reflects the experimental results?

A. There is more DAA in fried potatoes than in raw, uncooked potatoes.B. There is more AA in fried potatoes than in raw, uncooked potatoes.C. There is more total vitamin C content in fried than in raw, uncooked potatoes.D. There is more AA in fried potatoes compared with uncooked potatoes, but only if

they are fried at 140oC or lower.

Answer: A

Description: The concentration of DAA in uncooked potatoes is 7.4 mg/100 grams of dry matter. In the four examples of frying, the DAA concentrations ranged from 29.7 to42.8, clearly indicating an increase.

Q. 7 If some potatoes were boiled in a 10% NaCl solution, and there were 300 grams of potato left, how many milligrams of vitamin C would there be?

A. 14.7B. 17.4C. 26.7D. 44.1

Answer: D

Description: After potatoes are boiled in a 10% NaCl solution, there are 14.7 milligramsof vitamin C per 100 grams of potato. Therefore, if there were 300 grams of potato, therewould be 14.7 * 3 = 44.1 milligrams of vitamin C.


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Q. 8 What concentration of ascorbic acid would you expect to see if potatoes were friedfor 45 minutes at 140oC?

A. 0 mg/100 gramsB. 7.3 mg/100 grams

C. 20.6 mg/100 gramsD. There is insufficient evidence to answer

Answer: A

Description: Although this question does break our rule concerning extrapolation, it isreasonable to infer from the given data that the ascorbic acid will be depleted. The datashow a decreasing concentration until there is 0 mg/100 grams after 30 minutes of frying.Unless there is some process by which AA is suddenly created as the frying proceeds, theconcentration will stay at zero.

Q. 9

If a potato is boiled in water with 5% NaCl, about what percentage of the originalvitamin C will be lost?

A. 31%B. 36%C. 52%D. 64%

Answer: D

Description: The table tells us how much Vitamin C will be retained. The concentrationwill be about 18 mg/100 grams of potato after cooking in a 5% NaCl solution, whereasfor the uncooked potato, the vitamin C content is 50 mg per 100 grams. That means about36% of the Vitamin C is retained. But this means that about 64% is lost.

Q. 10 All of the following explanations could account for the different pattern of DAAretention when comparing frying and boiling except:

A. When immersed in water, the AA is less likely to oxydize to become DAA.B. When immersed in water, DAA is hydrolized to become 2,3-diketogluconic acid.

C. Frying increases the likelihood for DAA to break down into other compounds.D. Frying dries out the potatoes and reduces the tendency for DAA to hydrolyze.

Answer: C

Description: If it were true that frying increased the likelihood of DAA breaking downinto other compounds, then we would expect to see less DAA when frying than when

boiling. But the opposite is true, so this explanation must be false.


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it jee 2006 sample question paper

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