The volume of an ideal gas is held constant. Determine the ratio P2/P1 of the final pressure to the initial pressure when the temperature of the gas rises (a) from 39 to 78 K (b) from 39.0 to 53.8 oC.

(a) P₂ / P₁ = 2 / 1

(b) P₂ / P₁ = 17.93 / 13

At constant volume, the pressure (P) of an ideal gas is directly proportional to its temperature (T) as stated by Joseph Gay-Lussac. i.e

P ∝ T

=> P = KT

=> P / T = K

=> (P₁ / T₁) = (P₂ / T₂) = K

=> (P₁ / P₂) = (T₁ / T₂) = K        

=> (P₂ / P₁) = (T₂ / T₁) = K         (i)

Where;

P₁ and P₂ are the initial and final pressures of the gas.

T₁ and T₂ are the initial and final temperatures of the gas.

(a) if temperature rises from 39 to 78 K;

This implies that;

T₁ = 39 K

T₂ = 78 K

Substitute these values into equation (i) as follows;

=> (P₂ / P₁) = (78 / 39)

=> (P₂ / P₁) = (26 / 13)

=> (P₂ / P₁) = (2 / 1)

Therefore, the ratio P₂ / P₁ = 2 / 1

(b) if temperature rises from 39.0 to 53.8 K;

This implies that;

T₁ = 39.0 K

T₂ = 53.8 K

Substitute these values into equation (i) as follows;

=> (P₂ / P₁) = (53.8 / 39)

=> (P₂ / P₁) = (17.93 / 13)

Therefore, the ratio P₂ / P₁ = 17.93 / 13

i agree kid

explanation:

a

force tends to be low and long diustance masses are high