A sample of gas at 10.0 ATM and 5.0 °C increases in tempature of 35 °C. If the volume is unchanged, what is the new pressure?

11.08 atm

Explanation:

From the question given above, the following data were obtained:

Initial pressure (P₁) = 10 atm

Initial temperature (T₁) = 5 °C

Final temperature (T₂) = 35 °C

Final pressure (P₂) =?

Volume = constant

Next, we shall convert celsius temperature to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

Initial temperature (T₁) = 5 °C

Initial temperature (T₁) = 5 °C + 273

Initial temperature (T₁) = 278 K

Final temperature (T₂) = 35 °C

Final temperature (T₂) = 35 °C + 273

Final temperature (T₂) = 308 K

Finally, we shall determine the final pressure of the gas as follow:

Initial pressure (P₁) = 10 atm

Initial temperature (T₁) = 278 K

Final temperature (T₂) = 308 K

Final pressure (P₂) =?

P₁/T₁ = P₂/T₂

10/278 = P₂/308

Cross multiply

278 × P₂ = 10 × 308

278 × P₂ = 3080

Divide both side by 278

P₂ = 3080 / 278

P₂ = 11.08 atm

Therefore, the final pressure of the gas is 11.08 atm

the answer for sodium is 0.186

the answer for rubidium is 0.251