Question

The change in internal energy and temperature for 2 mole of ideal gas is +300 J and +5 K respectively. The molar heat capacity (in JK-'mol-l) at constant volume will be

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Answer 1

Given that:-

Pext.=1atm

P1=5atm

P2=2atm

T1=300K

T2=T(say)=?

n=2

CV=25R

For an adiabatic process,

W=nCVdT=−PΔV

nCV(T2−T1)=−Pext(V2−V1)

⇒2×25R(T−300)=−1(P2nRT2−P1nRT1)

⇒5R(T−300)=−2R(2T−5300)

⇒5T−1500=−T+120

⇒6T=1620⇒T=270K

∴T2=T=270K

∴W=nCVdT=2×25R×(270−300)=−1247.1J

⇒q=0

Δ

∴ΔH=ΔU+nRΔT

⇒ΔH=−1247.1+2×8.314×(270−300)=−1745.94J

Answer 2

The amount of heat needed to elevate 1 mole of a substance by 1 degree Kelvin is known as the molar heat capacity. Similar to specific heat capacity, molar heat capacity measures per mole rather than per gramme of substance.

Given

Number of moles, n = 2

Change in internal energy, ΔU = +300 J

Change in temperature, ΔT = +5 K

To find

molar heat capacity, $C_{v}$ = x $JK^{-1} mol^{-1}$ = ?

The relation between internal energy change and heat capacity.

ΔU = n$C_{v}$.ΔT

$C_{v}$ = ΔU / (n. ΔT)

Substitute the values.

$C_{v}$ = +300 J / (2 mol * 5 K)

=> 30 $JK^{-1} mol^{-1}$

Thus the correct answer is 30 J/ mol k.

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