Question

At a given temperature the pressure of a gas reduces to 75% of its initial value and the volume
increases by 40% of its initial value. Find this temperature if the initial temperature was – 10°C.
[3.15°C]

Answer 1

Question :

• At a given temperature the pressure of a gas is 75% of its initial value and the volume  increases by 40% of its initial value. Find this temperature if the initial temperature was – 10°C.

Given :

• Let the initial pressure be P and initial volume be V
• Initial temperature (T) = - 10 °C = 273 - 10 = 263 K
• No of moles (n) remains constant .

• Final pressure = P' =  0.75P = 0.75 P =
• Final volume = V' = V + 0.40 V = 1.4 V

Solution :

Initially ,

Ideal gas equation ,

PV = nRT

• P = pressure (atm )
• V = volume ( L)
• n = moles ( mol )
• R =I deal gas constant  (0.0821 L .atm / K. mol )
• T = Temperature ( K)

⇒ nR = PV / T ... (1)

Now ,

⇒ P' V' = n R T'

On rearranging ,

⇒ n R= P' V' / T' ... ..(2)

Equating equation (1) and (2) we get ,

⇒ PV / T = P' V' /T'

On rearranging ,

⇒ T ' = P' x V' x T / P x V

⇒ T' = 0.75P x 1.4 V x 263 / P x V

⇒ T' = 0.75 x 1.4 x 263

⇒ T' = 276.15  K

Conversion from K into °C

⇒ T ' = 276.15 - 273

⇒ T' = 3.15 °C.

The final temperature of the gas is 3.15 °C.