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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
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
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
An ideal monoatomic gas is taken round that cycle ABCDA as shown in the pV diagram (see figure). The work done during the cycle is
Work done in a cyclic process = area under the graph in pV diagram
Area = AB × BC = (2p – p) × (2V – V) = pV
∴ So, work done = pV
The correct answer is: pV
In a given process of an ideal gas, dW = 0 and dQ < 0. The for the gas
Select one:
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
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
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
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:
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 viceversa.
Hence, graph will be rectangular hyperbola as shown in the above figure.
The correct answer is:
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:
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
The corresponding pV 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}
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:
Slope of adiabatic process at a given state (p, V, T) is more than the slope of isothermal process. The corresponding pV 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
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:
Heat lost by steam =Heat gained by
(water +calorimeter)
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