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QUESTION: 1

A vessel contains 1 mole of O_{2} gas (molar mass 32) at a temperature T. The pressure of the gas is p. An identical vessels containing one mole of the gas (molar mass 4) at a temperature 2T has a pressure of

Solution:

QUESTION: 2

A gas mixture consists of 2 moles of oxygen and 4 moles of argon at temperature T. Neglecting all vibrational modes, the total internal energy of the system is

Solution:

Internal energy of n moles of an ideal gas at temperature T is given by

where, f = degrees of freedom

= 5 for O_{3} and 3 for Ar [∴ O_{2} is diatomic, Ar is monoatomic]

Hence,

The correct answer is: 11 RT

QUESTION: 3

An ideal monoatomic gas is taken round that cycle ABCDA as shown in the p-V diagram (see figure). The work done during the cycle is

Solution:

Work done in a cyclic process = area under the graph in p-V diagram

Area = AB × BC = (2p – p) × (2V – V) = pV

∴ So, work done = pV

The correct answer is: pV

QUESTION: 4

In a given process of an ideal gas, dW = 0 and dQ < 0. The for the gas

Select one:

Solution:

From first law of thermodynamics,

dQ = dU + dW

dQ = dU if dW = 0

Since, dQ < 0

Therefore, or temperature will decrease

The correct answer is: the temperature will decrease

QUESTION: 5

70 cal of heat are required to raise the temperature of 2 moles of an ideal diatomic gas at constant pressure from 30ºC to 35°C. The amount of heat required (in calorie) to raise the temperature of the same gas through the same range (30°C to 35°C) at constant volume is

Solution:

We know that heat required to raise the temperature at constant pressure is C_{p}.

So,

Similarly, at constant volume C_{V}

or

The correct answer is: 50

QUESTION: 6

An ideal gas is initially at temperature T and volume V. Its volume is increased by ΔV due to an increase in temperature ΔT pressure remaining constant. The quantity varies with temperature as

Select one:

Solution:

For an ideal gas : pV = nRT

For p = constant

Therefore,δ is inversely proportional to temperature T i.e., when T increases, δ decreases and vice-versa.

Hence, graph will be rectangular hyperbola as shown in the above figure.

The correct answer is:

QUESTION: 7

When an ideal diatomic gas is heated at constant pressure, the fraction of the heat energy supplied which increases the internal energy of the gas is

Select one:

Solution:

QUESTION: 8

Starting with the same initial conditions, an ideal gas expands from volume V_{1} to V_{2} in three different ways, the work done by the gas is W_{1} if the process is purely isothermal, W_{2} if purely isobaric and W_{3} if purely adiabatic, then

Solution:

The corresponding p-V graph (also called indicator diagram) in three different process will be as shown :

Area under the graph gives the work done by the gas.

(Area)_{2} > (Area)_{1} > (Area)_{3}

∴ W_{2} > W_{1} > W_{3}

The correct answer is:

W_{2} > W_{1} > W_{3}

QUESTION: 9

An ideal gas expands isothermally from a volume V_{1} to V_{2} and then compressed to original volumes V_{1} adiabatically. Initial pressure is p_{1} and final pressure is p_{3. }The total work done is W. Then,

Select one:

Solution:

Slope of adiabatic process at a given state (p, V, T) is more than the slope of isothermal process. The corresponding p-V graph for the two process is as shown in figure.

In the graph, AB is isothermal and BC is adiabatic.

W_{AB} is positive (as volume is increasing)

and W_{BC} is negative (as volume is decreasing)

as area under p–V graph gives the work done.

Hence,

From the graph itself, it is clear that p_{3} > p_{1}

The correct answer is: p_{3 }> p_{1} , W < 0

QUESTION: 10

Steam at 100°C is passed into 1.1 kg of water contained in a calorimeter of water equivalent 0.02 kg at 15°C till the temperature of the calorimeter and its contents rises to 80°C. The mass of the steam condensed (in kg) is

Select one:

Solution:

Heat lost by steam =Heat gained by

(water +calorimeter)

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