Question

Consider a balloon that has a volume V. It contains n moles of gas, it has an internal pressure of P, and its temperature is T. If the balloon is heated to a temperature of 15.5T while it is placed under a high pressure of 15.5P, how does the volume of the balloon change?

Answer 1

b.) it stays the same

Answer 2

Answer: The temperature will not change and will remain T.

Step-by-step explanation: Accoding to ideal gas law:

`PV=nRT`

where,

P = pressure of the gas

V = volume of the gas

T = temperature of the gas

n = number of moles of gas

R = Gas constant = 0.0821 Latm/moleK

`(P_1V_1)/(T_1)=(P_2V_2)/(T_2)`

`P_1`= initial pressure

`V_1`= initial volume

`T_1`= initial temperature

`P_2`= final pressure

`V_2`= final volume

`T_2`= final temperature

Putting in the values:

`(PV)/(T)=(15.5PV_2)/(15.5T)`

`T_2=T`