Chemical Engineering - Chemical Engineering Thermodynamics - Discussion

The Answer is 327K and not 327 degCelsius.

As we know for the ideal gas follows the following equation.

PV1=nRT1----- (1).

PV2=nRT2----- (2).

In the problem, given conditions are Pressure is constant & Volume is getting doubled.

So the two equations becomes.

1. V1=27nR.

2. 2V1=nRT2.

After taking ratio, calculated for T2 gives 54 degC (327K).

The answer is 327deg celsius
As we know for the ideal gas follows the following equation.

PV1=nRT1----- (1).

PV12=nRT2----- (2).

Try to assume that there is one mole of the ideal gas, 1 atm pressure and 0.08205 L-atm/mole-k as R
Use PV=nRT.

V=nRT/P
V=(1 mole)(0.08205 L-atm/mole-K)(27+273k)/(1 atm)
As you calculate, the result is 24.615 L

Now,the problem states that the volume has been doubled,therefore we must multiply the V=24.615 by 2 and the result is 49.23L
Let us go back to PV=nRT.

Substitute 49.23 to the value of V.
T=PV/nR
T=(1 atm)(49.23 L)/(1mole)(0.08205L-atm/mole-K)
T=600K

As the choices given are in degrees Celsius we must subtract 600K with 273 in equation:
T=600k-273
T=327 degrees Celsius.

PV = nR(27+273.15) ---> 1.

P(2V) = nRT --->2.

Dividing 1 by 2.

PV = nR(27+273.15).

---------------.

P(2V) = nRT.

P, V, n, and are cancels out.

(1/2) = (300.15 K)/T;

T = 600.3-273.15 = 327.15 degC.

Everyone should note that when using equations of state, we use absolute values for temperature and pressure.

Temperature given -27 degC.

Convert in kelvin-27+273 = 300k.

Given isobaric process so (v1/v2 = T1/T2).

Given V2 = 2V1. So 1/2 = 300/x.

From this X = 600k or 600-273 = 327 degC.

Your answer is best @Susmit.

Awesome explanation @Sushmit.

Here is the easy way to solve this: by using Charles law:

T1/V1 = T2/V2.
27 + 273/1 = T2/2.
T2 = 300K(2)/1.
T2 = 600k since the answer is in celcius.
T2 = 600 - 273 = 327 celcius.

327K not C.

Sumit has excellent concept. Good.