Question

A $ 15g $ mass of nitrogen gas is enclosed in a vessel at temperature $ {{27}^{\circ }}C $ . Amount of heat transferred to the gas, so that rms velocity of molecules is doubled, is about:
  $ \text{A}\text{. }10KJ $
  $ \text{B}\text{. }0.9KJ $
  $ \text{B}\text{. 6}KJ $
  $ \text{B}\text{. 14}KJ $

Answer 1

Hint: We will calculate the amount of heat transferred to the gas by applying the equation for heat required for raising the temperature of the system by $ 1K $ at constant volume. RMS velocity of gas is directly proportional to the square root of temperature.

Formula used:
For constant pressure, $ Q=n{{C}_{P}}dt $
For constant volume, $ Q=n{{C}_{V}}\Delta T $

Complete step-by-step answer:
We are given that the gas is enclosed in a vessel; it means that the gas undergoes an isochoric process.
An isochoric process is the thermodynamic process in which volume of the gas is held constant; meaning the work done by the gas is zero. Any heat energy transferred to the system externally is absorbed as the internal energy of the system or the gas.
Heat capacity of a gas is expressed as the amount of heat required to raise the temperature of a certain amount of gas by $ 1K $ .
The value of Heat capacity depends on whether the heat is added at constant pressure or at constant volume.
Expression for heat required to increase the temperature of system by $ 1K $ at constant pressure,
  $ Q=n{{C}_{P}}dt $
Expression for heat required to increase the temperature of system by $ 1K $ at constant volume,
  $ Q=n{{C}_{V}}dt $
As the gas is enclosed in vessel, it is undergoing isochoric process, so,
  $ Q=n{{C}_{V}}\Delta T $
Number of moles of Nitrogen gas, $ n=\dfrac{15}{28} $
For diatomic gases, $ {{C}_{V}}=\dfrac{5}{2}R $
RMS velocities of gas molecules is directly proportional to the square root of temperature
As the RMS velocity of gas molecules is doubled, it means that the temperature of gas has become four times the initial temperature
  $ T'=4T $
Temperature $ T $ is given as $ {{27}^{\circ }}C $ , or $ 300K $
It gives,
  $ \begin{align}
  & \Delta T=4T-T \\
 & \Delta T=3T=3\times 300 \\
 & \Delta T=900 \\
\end{align} $
Also, $ R=8.314J $
We have, $ Q=n{{C}_{V}}\Delta T $
  $ \begin{align}
  & Q=\dfrac{15}{28}\times \dfrac{5\times R}{2}\times 900 \\
 & Q=\dfrac{15}{28}\times \dfrac{5\times 8.314}{2}\times 900=1002.13\approx 10,000 \\
\end{align} $
  $ Q=10000J $
The amount of heat transferred to the gas is $ 10KJ $
Hence, the correct option is A.

Note: In the above question, we were given that the vessel is kept closed while transferring heat to the gas, it means the volume of the gas will remain constant throughout the process. Therefore we used the equation for heat required to raise the temperature of the system by $ 1K $ at constant volume. Also, while applying the equation, temperature or the change in temperature is always taken in kelvin. $\text{Temperature in Kelvin = Temperature in Celsius + 273} $.