Question

Question 8
A gas at 30 À°C is heated at
constant pressure till its volume is
doubled. Find the final temperature
of the gas.
151.5 degree kelvin
0.0066 degree kelvin
606 degree kelvin
60 degree kelvin​

Answer 1

Answer:

your answer

Step-by-step explanation:

Use the fact that V is proportional to T when P is constant

V1= KT1

V2 = KT2

KT2/KT1 = V2/ V1

T2= T1( V2/V1) = (273+30)(2)= 303×2 = 606K = (606–273)°C = 333°C

Answer 2

The final temperature of the gas is 606 Kelvin.

To solve this problem, we can use the combined gas law, which relates the initial and final temperatures, pressures, and volumes of the gas:

(P1/T1) = (P2/T2) * (V2/V1), where P1 and T1 are the initial pressure and temperature, P2 is the constant pressure during heating, V1 is the initial volume, V2 is the final volume (which is twice the initial volume), and we want to find T2, the final temperature.

We can simplify the equation by multiplying both sides by T2 and dividing both sides by P2:

T2 = (P2 * V2 * T1) / (P1 * V1)

Substituting the given values into this equation, we get:

T2 = (P2 * V2 * T1) / (P1 * V1)

= (1 * 2 * V1 * (30 + 273)) / (1 * V1 * (30 + 273))

= 2 * (30 + 273)

= 606 K

Therefore, the final temperature of the gas is 606 Kelvin. Answer: 606 degree kelvin.

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