Question

1
mole of Helium gas is contained in a
container at S.T.P. The heat energy needed
to double the pressure of the gas, keeping
the volume constant (heat capacity of the
gas=3Jg-'K-') is
1) 3276J 2) 1638J3) 819J 4) 409.5J​

Answer 1

Answer:

A.1638 J

Explanation:

Here, n=

2

1

,c

v

=3Jg

−1

K

−1

, M=4g mol

−1

∴C

v

=M

C

V

4×3=12J mol

−1

K

−1

At constant volume P∝T.

∴

P

1

P

2

=

T

1

T

2

=2,T

2

=2T

1

Rise in temperature ΔT=T

2

−T

1

=2T

1

−T

1

=T

1

=273K

Heat required, ΔQ=NC

V

ΔT=

2

1

×12×273=1638J

Answer 2

Answer:

thanks for free points

Explanation:

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