One mole of a monoatomic ideal gas undergoes the process \(A\rightarrow B\) in the given \(P-V\) diagram. The molar heat capacity for this process is:

1. \(\frac{3R}{2}\)
2. \(\frac{13R}{6}\)
3. \(\frac{5R}{2}\)
4. \(2R\)

P-V diagram of a diatomic gas is straight line passing through origin. The molar heat capacity of the gas in the process will be

1. 4R

2. 2.5 R

3. 3R

4. $\frac{4\mathrm{R}}{3}$

The pressure of a monoatomic gas increases linearly from $4\times {10}^5$ N/m² to $8\times {10}^5$ N/m² when its volume increases from 0.2 m³ to 0.5 m³. Calculate  molar heat capacity of the gas [R = 8.31 J/mol k]  

1. 20.1 J/molK               

2. 17.14 J/molK

3. 18.14 J/molK                                   

20.14 J/molK

At ordinary temperatures, the molecules of a diatomic gas have only translational and rotational kinetic energies. At high temperatures, they may also have vibrational energy. As a result of this compared to lower temperatures, a diatomic gas at higher temperatures will have:

If the temperature of source & sink in heat engine is
at 1000 K & 500 K respectively, then efficiency
can be
(1) 20%

(2) 30%

(3) 50%

(4) All of these

In an adiabatic process, work done versus change of
temperature $∆$T is

The volume and temperature graph is given in the figure below. If pressures for the two processes are different, then which one,  of the following, is true?

n moles of an ideal gas is heated at constant pressure
from 50°C to 100°C, the increase in internal energy
of the gas is $\left(\frac{{\mathrm{C}}_{\mathrm{p}}}{{\mathrm{C}}_{\mathrm{v}}}=\mathrm{\gamma }<mspace\; width="1em"></mspace>\mathrm{and}<mspace\; width="1em"></mspace>\mathrm{R}=\mathrm{gas}<mspace\; width="1em"></mspace>\mathrm{constant}\right)$

1. $\frac{50<mspace\; width="1em"></mspace>\mathrm{nR}}{\mathrm{\gamma }-1}$

2. $\frac{100<mspace\; width="1em"></mspace>\mathrm{nR}}{\mathrm{\gamma }-1}$

3. $\frac{50<mspace\; width="1em"></mspace>\mathrm{n\gamma R}}{\mathrm{\gamma }-1}$

4. $\frac{25<mspace\; width="1em"></mspace>\mathrm{n\gamma R}}{\mathrm{\gamma }-1}$

The molar heat capacity C for an ideal gas going through a given process is given by C = a/T , where 'a' is a constant. If  $\mathrm{\gamma }=<mspace\; width="1em"></mspace>{\mathrm{C}}_{\mathrm{P}}/{\mathrm{C}}_{\mathrm{V}}$ , the work done by one mole of gas during heating from ${\mathrm{T}}_0$ to $\mathrm{\eta }<mspace\; width="1em"></mspace>{\mathrm{T}}_0$ through the given process will be:

1.  $\frac{1}{\mathrm{a}}<mspace\; width="1em"></mspace>\ln <mspace\; width="1em"></mspace>\mathrm{\eta }$

2.  $\mathrm{a}<mspace\; width="1em"></mspace>\ln <mspace\; width="1em"></mspace>\mathrm{\eta }-<mspace\; width="1em"></mspace>\left(\frac{\mathrm{\eta }-1}{\mathrm{\gamma }-1}\right)<mspace\; width="1em"></mspace>{\mathrm{RT}}_0$

3.  $\mathrm{a}<mspace\; width="1em"></mspace>\ln <mspace\; width="1em"></mspace>\mathrm{\eta }-\left(\mathrm{\gamma }-\mathrm{A}\right)<mspace\; width="1em"></mspace>{\mathrm{RT}}_0$

4.  none of these

An ideal gas expands according to the law ${\mathrm{PV}}^2$ = const. The molar heat capacity C is : 

1. ${\mathrm{C}}_{\mathrm{V}}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>\mathrm{R}$ 

2. ${\mathrm{C}}_{\mathrm{V}}<mspace\; width="1em"></mspace>–<mspace\; width="1em"></mspace>\mathrm{R}$ 

3. ${\mathrm{C}}_{\mathrm{V}}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>2\mathrm{R}$

4. ${\mathrm{C}}_{\mathrm{V}}<mspace\; width="1em"></mspace>–<mspace\; width="1em"></mspace>3\mathrm{R}$