Chemical Engineering - Chemical Engineering Thermodynamics - Discussion
Discussion Forum : Chemical Engineering Thermodynamics - Section 1 (Q.No. 20)
20.
Keeping the pressure constant, to double the volume of a given mass of an ideal gas at 27°C, the temperature should be raised to __________ °C.
Discussion:
12 comments Page 2 of 2.
Judy said:
1 decade ago
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.
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.
Swaminathan said:
1 decade ago
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).
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).
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