Step 1: How rms speed relates to temperature
The root mean square (rms) speed of a gas,
Vrms, increases as the square root of its temperature (
T). So,
Vrms∝√T.
Step 2: Setting up the ratio
Since
Vrms2∝T, we can write:
[]2=Step 3: Substituting given values
The rms speed increases from
v to
√3v. So:
()2=This simplifies to
=, so
T2=3T1.
Step 4: Calculating heat required
The heat
(Q) needed to raise the temperature is:
Q=nCV∆TFor 4 moles of a diatomic gas,
CV=R, so:
Q=4×R(T2−T1)We know
T2=3T1, so
∆T=3T1−T1=2T1.
Step 5: Plugging in the numbers
We are given
Q=83.1kJ=83,100J and
R=8.31Jmol−1K−1.
83,100=4××8.31×2T14××2=20, so:
83,100=20×8.31×T120×8.31=166.2, so: 83,100=166.2×T1T1==500KStep 6: Converting to Celsius
T1 in Celsius =500−273=227∘C