Work done by 0.1 mole of a gas at 27° C. To double its volume at constant pressure, is: (R=2 cal / mol K )

Question

Work done by 0.1 mole of a gas at 27° C. To double its volume at constant pressure, is: (R=2 cal / mol K ) (A) 546 cal (B) 60 cal (C) 600 cal (D) 54 cal

Tardigrade Admin · Accepted Answer

Using ideal gas equation P V=μ R T ...(i) where, P is pressure, V is volume, R is gas constant and T is temperature. Also work done is W ≡ P Δ V =P(2 V-V)=P V ...(ii) Equating Eqs. (i) and (ii), we get W =P V=μ R T ∴ W =0.1 × 2 × 300=60 cal.

Q. Work done by $0.1$ mole of a gas at $2 7^{\circ} C$ . To double its volume at constant pressure, is : $(R = 2 c a l / m o l K)$

546 cal

60 cal

600 cal

54 cal

Using ideal gas equation
$P V = μ RT$ ...(i)
where, $P$ is pressure, $V$ is volume, $R$ is gas constant and $T$ is temperature.
Also work done is
$W \equiv P Δ V$
$= P (2 V - V) = P V$ ...(ii)
Equating Eqs. (i) and (ii), we get
$W = P V = μ RT$
$∴ W = 0.1 \times 2 \times 300 = 60 c a l .$

Q. Work done by 0.1 mole of a gas at 27∘C. To double its volume at constant pressure, is : (R=2cal/molK)