Question

If the temperature of a gas is increased from ${27}^{\circ }$ C to ${927}^{\circ }$C, the root mean square speed of its molecules

• A. becomes half
  
• B. is doubled
  
• C. becomes 4 times
  
• D. remains unchanged

Answer 1

Answer: (B)

. is doubled.

Step 1, Given

Initial temperature = ${27}^{0}C$

Final temperature = ${927}^{0}C$

Step 2,

The root mean square speed of a gas is given by

$V_{rms}=\sqrt{\frac{3kT}{m}}$
i.e. $V_{rms}\propto \sqrt{T}$.

Initial temperature $T_1=27+273=300K.$

Final temperature \( T_2=927+273=1200K.\2)

$\Rightarrow \frac{V_{rms2}}{V_{rms1}}=\sqrt{\frac{T_2}{T_1}}=\sqrt{\frac{1200}{300}}=2$

$V_{rms2}=2V_{rms1}$

Thus the root mean square speed is doubled.

Hence the correct choice is (A).