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 1 of 2.
Bhatt said:
2 years ago
No @Alya.
It's ° C.
It's ° C.
(1)
Sandip said:
5 years ago
For an ideal gas PV = nRT
So, at constant pressure-volume is directly proportional to temperature so to double the volume the temperature also has to be doubled, so the temperature has to be 327°C.
So, at constant pressure-volume is directly proportional to temperature so to double the volume the temperature also has to be doubled, so the temperature has to be 327°C.
Akil Suhail said:
6 years ago
Sumit has excellent concept. Good.
Alyaa said:
7 years ago
327K not C.
(1)
ImeldaDante said:
9 years ago
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.
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.
(20)
Harshal said:
9 years ago
Awesome explanation @Sushmit.
ANMOL said:
9 years ago
Your answer is best @Susmit.
Susmit said:
10 years ago
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.
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.
(10)
Remington Salaya said:
1 decade ago
Everyone should note that when using equations of state, we use absolute values for temperature and pressure.
Remington Salaya said:
1 decade ago
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.
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.
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