Question

One gram molecule of oxygen is heated at constant pressure from 0°C. What amount of heat should be imparted to the gas to double it's volume? Given cp = 0.218 cal g-1 0 C-1.
Class +1 ( Physics )
​

Answer 1

Answer:

ʏᴇs ᴏғᴄᴏᴜʀsᴇ ɪ ᴄᴀɴ ʙᴇ ᴜʀ ғʀɴᴅ

ᴍʏsᴇʟғ ɢᴀʀɢɪ

ᴄʟᴀss 7

ғʀᴏᴍ ᴍᴀʜᴀʀᴀsʜᴛʀᴀ

ᴜʀs ɪɴᴛʀᴏ ᴘʟᴇᴀsᴇ?

Explanation:

Given info : one gram molecule of oxygen is heated at constant pressure from 0° C.

To find : what amount of heat should be imparted to the had to double its volume ?

solution : one gram molecule means one mole.

so no of moles of oxygen gas , n = 1

work done , W = \int\limits^{2V}_{V}{p}\,dV

V

∫

2V

pdV = PV = nRT₀ = 1 × R T₀ = RT₀

internal energy change, ∆U = \int\limits^{2V}_{V}C_vdT

V

∫

2V

C

v

dT

= \int\limits^{2V}_V{\frac{R}{\gamma-1}}\,dT

V

∫

2V

γ−1

R

dT

we know, pdV = nRdT

so, \int\limits^{2V}_V{\frac{P}{\gamma-1}}\,dV

V

∫

2V

γ−1

P

dV

= \frac{pV}{\gamma-1}=\frac{RT_0}{\gamma-1}

γ−1

pV

=

γ−1

RT

0

now from 1st law of thermodynamics,

Q = W + ∆U

⇒Q = RT₀ + RT₀/(γ - 1)

for oxygen (diatomic molecule),

γ = 1.4

= 25/3 × 273 + 25/3 × 273/(1.4 - 1) J

= 2275 + 2275/0.4

= 2275 + 5687.5

= 7962.5 J

Therefore heat required to the gas to double its volume is 7962.5 J

ʜᴏᴘᴇ ᴛʜɪs ᴡɪʟʟ ʜᴇʟᴘ ᴜ ᴅᴇᴀʀ..

Answer 2

Answer:

ΔW=∫2VVpdV=p0V=RT0

ΔU=∫CvdT=∫2VVRγ−1dT

At constant pressure fixed at

p0dV=RdT(∵pV=RT)

∴ΔU=1γ−1∫2VVp0dV=p0Vγ−1=RT0γ−1

∴ΔQ=ΔU+ΔW=RT0γ−1+RT0=γRT0γ−1

Since oxygen is diatomic, γ=1.4

∴ΔQ=1.4×8.3×2731.4−1=7931J

Explanation:

hope its helpful .

hello i,m good what about u??