Question

One mole of an ideal diatomic gas
absorbs 1831.4J of heat and does
1000 of work at 27°C. What is the
change in temperature observed ?
1) 25°C
2) 40°C
3) 37°C
4) 67°C​

Answer 1

Explanation:

One mole of an ideal diatomic gas

absorbs 1831.4J of heat and does

1000 of work at 27°C. What is the

change in temperature observed ?

1) 25°C

2) 40°C

3) 37°C✓✓✓✓

4) 67°C

Answer 2

Answer:

  T2 = 40

Explanation:

Ideal gas equations are PV = n RT

And change in enthalpies is given by ∆U=q-W

Now we are given that;

∆U = q - W  

= 1831.4 - 1000  

= 831.4 J  

(T2 - T1) × 5/2 × R  = ∆U

  ∆U   =  (T2 - 300) × 5/2 × 8.314  

   T2  = 831.4 × 2/ 8.314 × 5  

  T2 = 40