Question

An ideal gas occupies a volume of 2 m³ at a pressure of 3 × 10⁶ Pa. The energy of the gas is:
(A) 9 × 10⁶J (B) 6 × 10⁴J
(C) 10⁸J (D) 3 × 10²J

Answer 1

Answer:

Translational Kinetic Energy of a Mono-atomic gas is given as:

→ 3/2 ( PV ) or 3/2 (nRT )   [ Since PV = nRT ]

(Degree of Freedom for a Mono-atomic gas is 3.)

According to the question,

• Pressure ( P ) = 3 × 10⁶ Pa
• Volume ( V ) = 2 m³

Substituting the values we get:

→ Energy = 3/2 × 3 × 10⁶ Pa × 2 m³

→ Energy = 9 × 10⁶ Pa.m³ or 9 × 10⁶ J

Hence option (A) is the correct answer.

Answer 2

QUESTION :-

An ideal gas occupies a volume of 2 m³ at a pressure of 3 × 10⁶ Pa. The energy of the gas is:

(A) 9 × 10⁶J (B) 6 × 10⁴J

(C) 10⁸J (D) 3 × 10²J

SOLUTION :-

$\star\rm{k.e \: of \: a \: gas \:due \:to \:translation}$ $\rm{per \:mole ,}$

$\implies \rm \: e = \frac{3}{2}pv \\ \: \rm \: {where, }\\ \implies \rm \: r \: = universal \: gas \: constant. \\ \implies \rm \: t \: = temperature \: of \: gas \\ \rm \: now, \\ \implies \rm \: p \: = 3 \times 10 {}^{6} \: and \: \: \: v \: = 2m {}^{3} \: \\ \implies \rm \:let \: gas \: is \:monoatomic\: \: \: \: .°. \: f \: = 3 \: \: \\ \implies \rm \: {\boxed{e = \frac{f}{2}nRT=\frac{f}{2}pv}} \: \longrightarrow \: \red \: {formula} \\ \implies \rm \: .°. \: e = \frac{3}{2} \times 3 \times 10 {}^{6} \times 2 \\ \implies \rm \: .°. \: e = \frac{3}{ \cancel2} \times 3 \times 10 {}^{6} \times \cancel2 \\ \implies \rm \: .°. \: e =3 \times 3 \times 10 {}^{6} \\ \implies \rm \: { \boxed{\boxed{.°. \: e =9 \times 10 {}^{6} J}}} \\ \\ \longrightarrow \rm \huge \underline{\: option \: (a) \: }$