Two Carnot engines x and y are working between the same source temperature \(T_1\) and the same sink temperature \(T_2\)${\mathrm{}}_{}$. If the temperature of the source in Carnot engine x is increased by \(\Delta T\), and in the Carnot engine y, the temperature of the sink is increased by\(\Delta T\), then the efficiency of x and y becomes \(\eta_\mathrm x\)${\mathrm{}}_{}$ and\(\eta_\mathrm y\)${\mathrm{}}_{}$. Then:

1.	\(\eta_{\mathrm{x}}=\eta_{\mathrm{y}}\)
2.	\(\eta_{\mathrm{x}}<\eta_{\mathrm{y}}\)
3.	\(\eta_{\mathrm{x}}>\eta_{\mathrm{y}}\)
4.	The relation between \(\eta_{\mathrm{x}}\) and \(\eta_{\mathrm{y}}\) depends on the nature of the working substance

In the P-V graph shown for an ideal diatomic gas, the change in the internal energy is:
                                
1. \(\frac{3}{2}P(V_2-V_1)\)
2. \(\frac{5}{2}P(V_2-V_1)\)
3. \(\frac{3}{2}P(V_1-V_2)\)
4. \(\frac{7}{2}P(V_1-V_2)\)

The first law of thermodynamics is based on:

The efficiency of an ideal heat engine is less than 100% because of:

A heat engine operates between the temperatures of 300 K and 500 K. If it extracts 1200 J of heat energy from the source, then the maximum amount of work that can be done by the engine is:

1. 720 J

2. 520 J

3. 480 J

4. 200 J

If 3 moles of a monoatomic gas do 150 J of work when it expands isobarically, then a change in its internal energy will be:

In the case of free expansion, when a sample of gas expands suddenly, the change in internal energy of the gas will be:

1. Positive

2. Negative

3. Zero

4. May be positive or negative

The work done for the given process AB is:

1. 6$P_0{V}_0$

2. 5$P_0{V}_0$

3. 3$P_0{V}_0$

4. 2$P_0{V}_0$

The efficiency of a Carnot heat engine working between the temperatures \(27^{\circ}\mathrm{C}\)$$ and \(227^{\circ}\mathrm{C}\) is:
1. 0.1
2. 0.6
3. 0.2
4. 0.4

Work done during the given cycle is:
                               
1. 4$P_0{V}_0$

2. 2$P_0{V}_0$

3. $\frac{1}{2}$$P_0{V}_0$

4. $P_0{V}_0$