TG EAMCET 2 May 2025 Shift 2 Paper

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:
[‌

(Vrms)1

(Vrms)2

]2=‌

T1

T2

Step 3: Substituting given values
The rms speed increases from v to √3v. So:
(‌

v

√3v

)2=‌

T1

T2

This simplifies to ‌

1

3

=‌

T1

T2

, so T2=3T1.
Step 4: Calculating heat required
The heat (Q) needed to raise the temperature is:
Q=nCV∆T
For 4 moles of a diatomic gas, CV=‌

5

2

R, so:
Q=4×‌

5

2

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×‌

5

2

×8.31×2T1
4×‌

5

2

×2=20, so:
83,100=20×8.31×T1
20×8.31=166.2‌, so: ‌83,100=166.2×T1
T1=‌

83,100

166.2

=500K
Step 6: Converting to Celsius
T1‌ in Celsius ‌=500−273=227∘C