Question

Eight moles of a monoatomic ideal gas undergoes a process given by PT2 = constant where p is density
and Tis temperature of gas in Kelvin. Then heat supplied to the gas to increase its temperature by 27°C
is (R = 25/3J/mol/K)
​

Answer 1

Answer:

Given,

P∝V

−2

n=2

T

1

=300K

T

2

=400K

PV

2

=const. . . . . .(1)

γ=2 (PV

γ

=constant)

The above equation shows that the process is adiabatic process,

The work done in adiabatic process in thermodynamics is given by

W=nR

γ−t

(T

1

−T

2

)

W=nR

2−1

(300−400)

W=−100nR=−2×100R

W=−200R

Answer 2

Concept introduction:

In varying pressure with temperattue there is a law called Boyle's law and it sates that the pressure (p) of a given amount of gas varies inversely with its volume (v) at a teperature with constant value; i.e., in equation form, pv = constant which is denotes as k.

Given:

Here it is given that $PT2 = constant$.

$P$ is the density.

To find:

We have to find that  heat supplied to the gas to increase its temperature by $27$$C$.

Solution:

According to the question,

$P$ $\alpha$ $V$

$T1=300K\\T2=300K$

$PV2=constant$

$W=(300-400)nR\\- > W=-100*2*R\\- > W=-200R$

Hence, the answer is $-200R$ $J$.