Question 20
Gas LawsThe pressure of two moles of an ideal gas at a temperature of 27°C and volume 10−2m3 is [R = 8.313 J mol−1K−1]
- A.4.99 × 105 Nm−2.
- B.9.80 × 103 Nm−2.
- C.4.98 × 103 Nm−2.
- D.9.80 × 105 Nm−2.
Show the answer and explanation
Answer: A (4.99 × 105 Nm−2.)
Explanation
PV = nRT, so P = 2 × 8.313 × 300 ÷ 10−2 ≈ 4.99 × 105 N m−2.