The volume of a balloon is 1 liter at 1 atm pressure and 0°C temperature.
What volume is the ballooon at 2 atm pressure and 273°C temperature?
I hope you may have learned about the Ideal gas equation(PV=nRT), where P=Pressure of the gas , V=Volume of the gas , n=amount of gas/substance(in moles) , R= Universal gas constant , T= Absolute Temperature of the gas
We use this equation for the calculations regarding real gasses,assuming they are ideal gasses.
So in this case, n is a constant and the product of PV is directly propotional to T(as R is a constant).
So (PV)/T =nR(and it is a constant)
So simply substitute values for the left side of the equation, in these two cases and equate them to find the required volume.
Hope this helps.
To find the volume of the balloon at different pressure and temperature, we can use the Ideal Gas Law equation, which states that the product of the pressure (P), volume (V), and temperature (T) of a gas is proportional to the number of moles (n) and the ideal gas constant (R):
PV = nRT
Where:
P = Pressure
V = Volume
T = Temperature
n = Number of moles of gas
R = Ideal gas constant
Since the number of moles remains constant, we can rewrite the equation as:
P1V1 / T1 = P2V2 / T2
Where:
P1 = Initial Pressure
V1 = Initial Volume
T1 = Initial Temperature
P2 = New Pressure
V2 = New Volume
T2 = New Temperature
Now, let's substitute the given values into the formula:
P1 = 1 atm
V1 = 1 liter
T1 = 0 °C (which is equivalent to 273 K, as 0 °C = 273 K)
P2 = 2 atm
T2 = 273 °C (which is equivalent to 546 K, as 273 °C = 546 K)
Plugging in these values, we get:
(1 atm)(1 liter) / (273 K) = (2 atm)(V2) / (546 K)
Now, let's solve for V2:
V2 = [(1 atm)(1 liter)(546 K)] / [(2 atm)(273 K)]
V2 = 3 liters
Therefore, the volume of the balloon at 2 atm pressure and 273 °C temperature is 3 liters.