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If the kinetic energy of one mole of an ideal gas is given as E=3/2 RT, (where R is the universal gas constant and T is the absolute temperature of the gas), the molar specific heat at constant pressure will be given as
  • a)
    2.5 R
  • b)
    1.5 R
  • c)
    0.5 R
  • d)
    0.1 R
Correct answer is option 'A'. Can you explain this answer?
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If the kinetic energy of one mole of an ideal gas is given as E=3/2 R...
Molar Specific Heat at Constant Pressure

Molar specific heat at constant pressure is defined as the amount of heat required to raise the temperature of one mole of a substance by 1 degree Celsius while keeping the pressure constant. It is denoted by Cp.

Formula for Molar Specific Heat at Constant Pressure

The formula for molar specific heat at constant pressure is given as:

Cp = (dH/dT)p

where dH is the change in enthalpy and dT is the change in temperature.

Relation between Enthalpy and Internal Energy

Enthalpy (H) is defined as the sum of internal energy (U) and the product of pressure (P) and volume (V). That is:

H = U + PV

Therefore, the change in enthalpy can be expressed as:

dH = dU + PdV + VdP

Since the process is at constant pressure, dP = 0, and therefore:

dH = dU + PdV

At constant pressure, the change in enthalpy is equal to the heat absorbed or released by the system. That is:

dH = q

where q is the heat absorbed or released by the system.

Therefore, the formula for molar specific heat at constant pressure can be expressed as:

Cp = (dq/dT)p

Since the process is at constant pressure, dq = dH. Therefore:

Cp = (dH/dT)p

Substituting the expression for dH, we get:

Cp = (dU/dT)p + P(dV/dT)p

Molar Specific Heat of an Ideal Gas

For an ideal gas, the internal energy depends only on the temperature and is given by the formula:

U = 3/2 RT

where R is the universal gas constant.

Therefore, the change in internal energy can be expressed as:

dU = 3/2 RdT

Since the process is at constant pressure, the volume changes according to the ideal gas law:

PV = nRT

where n is the number of moles of the gas.

Differentiating both sides with respect to temperature, we get:

V(dP/dT) + P(dV/dT) = nR

Since the process is at constant pressure, dP/dT = 0. Therefore:

(dV/dT)p = nR/P

Substituting these expressions in the formula for Cp, we get:

Cp = 3/2 R + 5/2 R = 2.5 R

Therefore, the molar specific heat at constant pressure for an ideal gas is 2.5 R, where R is the universal gas constant and depends on the units used.
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Direction: Read the case Study given below and answer an three of the questions that follow.The temperature-depth profile for the ocean water shows how the temperature decreases with the increasing depth. The profile shows a boundary region between the surface waters of the ocean and the deeper layers. The boundary usually begins around 100 - 400 m below the sea surface and extends several hundred of metres downward. This boundary region, from where there is a rapid decrease of temperature, is called the thermocline. About 90 per cent of the total volume of water is found below the thermocline in the deep ocean. In this zone, temperatures approach 0°C. The temperature structure of oceans over middle and low latitudes can be described as a three-layer system from surface to the bottom. The first layer represents the top layer of warm oceanic water and it is about 500 m thick with temperatures ranging between 20° and 25°C. This layer, within the tropical region, is present throughout the year but in mid latitudes it develops only during summer. The second layer called the thermocline layer lies below the first layer and is characterised by rapid decrease in temperature with increasing depth. The thermocline is 500-1,000 m thick. The third layer is very cold and extends upto the deep ocean floor. In the Arctic and Antarctic circles, the surface water temperatures are close to 0°C and so the temperature change with the depth is very slight. Here, only one layer of cold water exists, which extends from surface to deep ocean floor.Q. The __________ profile for the ocean water shows how the temperature decreases with the increasing depth.

Direction: Read the case Study given below and answer an three of the questions that follow.The temperature-depth profile for the ocean water shows how the temperature decreases with the increasing depth. The profile shows a boundary region between the surface waters of the ocean and the deeper layers. The boundary usually begins around 100 - 400 m below the sea surface and extends several hundred of metres downward. This boundary region, from where there is a rapid decrease of temperature, is called the thermocline. About 90 per cent of the total volume of water is found below the thermocline in the deep ocean. In this zone, temperatures approach 0°C. The temperature structure of oceans over middle and low latitudes can be described as a three-layer system from surface to the bottom. The first layer represents the top layer of warm oceanic water and it is about 500 m thick with temperatures ranging between 20° and 25°C. This layer, within the tropical region, is present throughout the year but in mid latitudes it develops only during summer. The second layer called the thermocline layer lies below the first layer and is characterised by rapid decrease in temperature with increasing depth. The thermocline is 500-1,000 m thick. The third layer is very cold and extends upto the deep ocean floor. In the Arctic and Antarctic circles, the surface water temperatures are close to 0°C and so the temperature change with the depth is very slight. Here, only one layer of cold water exists, which extends from surface to deep ocean floor.Q. Which layer is characterised by rapid decrease in temperature with increasing depth?

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If the kinetic energy of one mole of an ideal gas is given as E=3/2 RT, (where R is the universal gas constant and T is the absolute temperature of the gas), the molar specific heat at constant pressure will be given asa)2.5 Rb)1.5 Rc)0.5 Rd)0.1 RCorrect answer is option 'A'. Can you explain this answer?
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