One mole of diatomic gas having rotational modes only is kept in a cylinder with a piston system. The cross-section area of the cylinder is $4 \text{ cm}^2$. The gas is heated slowly to raise the temperature by $1.2^\circ\text{C}$ during which the piston moves by $25 \text{ mm}$. The amount of heat supplied to the gas is . . . . . . $J$. (Atmospheric pressure = $100 \text{ kPa}$,$R = 8.3 \text{ J/mol}\cdot\text{K}$) (Neglect mass of the piston)
- A$24.8$
- B$25$
- C$15.04$
- D$29.98$
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An ideal monoatomic gas is carried along the cycle $ABCDA$ as shown in the figure. The total heat absorbed during this process is (in $p_0 V_0$)
Match the following:
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The correct answer is:
| List-$I$ | List-$II$ |
| $i)$ Isothermal process | $a)$ $0$ |
| $ii)$ Isobaric process | $b)$ $\frac{1}{\gamma-1}[P_2 V_2 - P_1 V_1]$ |
| $iii)$ Isochoric process | $c)$ $\mu RT \ln(\frac{V_2}{V_1})$ |
| $iv)$ Adiabatic process | $d)$ $P(V_2 - V_1)$ |
The correct answer is:
Two cylinders $A$ and $B$ fitted with pistons contain equal amounts of an ideal diatomic gas at $300 \ K$. The piston of $A$ is free to move,while that of $B$ is held fixed. The same amount of heat is given to the gas in each cylinder. If the rise in temperature of the gas in $A$ is $30 \ K$,then the rise in temperature of the gas in $B$ is ..... $K$.
MediumAIIMS 2019
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