An ideal gas at 20 oC is heated isobarically. Find the temperature of the gas, when volume of gas getsdoubled

To find the temperature of the gas when the volume is doubled, we can use the ideal gas law equation:

PV = nRT

Where: P = pressure of the gas V = volume of the gas n = number of moles of gas R = ideal gas constant T = temperature of the gas

Since the gas is heated isobarically, the pressure remains constant. Therefore, we can rewrite the equation as:

V1/T1 = V2/T2

Where: V1 = initial volume of the gas T1 = initial temperature of the gas V2 = final volume of the gas (doubled) T2 = final temperature of the gas (to be found)

We are given that the initial temperature of the gas is 20 oC. However, we need to convert it to Kelvin since the ideal gas law requires temperature in Kelvin. The conversion formula is:

T(K) = T(°C) + 273.15

So, the initial temperature in Kelvin is:

T1 = 20 + 273.15 = 293.15 K

We are also given that the volume of the gas is doubled. Therefore, we can write:

V2 = 2 * V1

Now, we can substitute the values into the equation and solve for T2:

V1/T1 = V2/T2

V1/(20 + 273.15) = (2 * V1)/T2

Cross-multiplying, we get:

V1 * T2 = (2 * V1) * (20 + 273.15)

Simplifying further:

T2 = (2 * V1 * (20 + 273.15)) / V1

T2 = 2 * (20 + 273.15)

T2 = 2 * 293.15

T2 = 586.3 K

Therefore, the temperature of the gas when the volume is doubled is 586.3 K.