Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

$4.0 \text{ g}$ of a gas occupies $22.4 \text{ L}$ at NTP. The specific heat capacity of the gas at constant volume is $5.0 \text{ J K}^{-1}\text{mol}^{-1}$ . If the speed of sound in this gas at NTP is $952 \text{ ms}^{-1}$ , then the heat capacity at constant pressure is: (Take gas constant $R = 8.3 \text{ J K}^{-1}\text{mol}^{-1}$ )

$8.0 \text{ J K}^{-1}\text{mol}^{-1}$

$7.5 \text{ J K}^{-1}\text{mol}^{-1}$

$7.0 \text{ J K}^{-1}\text{mol}^{-1}$

$8.5 \text{ J K}^{-1}\text{mol}^{-1}$

Step-by-Step Solution

Find Molar Mass ( $M$ ): At NTP (Normal Temperature and Pressure, $T = 273 \text{ K}, P = 1 \text{ atm}$ ), $1 \text{ mole}$ of an ideal gas occupies $22.4 \text{ L}$ . Given that $4.0 \text{ g}$ of the gas occupies $22.4 \text{ L}$ , the molar mass of the gas is: $M = 4.0 \text{ g mol}^{-1} = 4.0 \times 10^{-3} \text{ kg mol}^{-1}$
Calculate the Ratio of Specific Heats ( $\gamma$ ): The speed of sound $v$ in a gas is given by Laplace's formula : $v = \sqrt{\frac{\gamma R T}{M}}$ Squaring both sides and solving for $\gamma$ : $\gamma = \frac{v^2 M}{R T}$ Substitute the given values ( $v = 952 \text{ ms}^{-1}$ , $M = 4.0 \times 10^{-3} \text{ kg mol}^{-1}$ , $R = 8.3 \text{ J K}^{-1}\text{mol}^{-1}$ , $T = 273 \text{ K}$ ): $\gamma = \frac{(952)^2 \times 4.0 \times 10^{-3}}{8.3 \times 273} = \frac{906304 \times 0.004}{2265.9} = \frac{3625.216}{2265.9} \approx 1.6$
Calculate Heat Capacity at Constant Pressure ( $C_p$ ): We know that the ratio of molar heat capacities is $\gamma = \frac{C_p}{C_v}$ . Given $C_v = 5.0 \text{ J K}^{-1}\text{mol}^{-1}$ , we can find $C_p$ : $C_p = \gamma \times C_v = 1.6 \times 5.0 = 8.0 \text{ J K}^{-1}\text{mol}^{-1}$

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