The pressure exerted by a non-reactive gaseous mixture of 6.4 g of methane and 8.8 g of carbon dioxide in a 10 L vessel at 27° C is kPa. (Round off to the Nearest Integer). [Assume gases are ideal, R =8.314 J mol -1 K -1 Atomic masses: C: 12.0 u , H: 1.0 u , O: 16.0 u ]

Question

The pressure exerted by a non-reactive gaseous mixture of 6.4 g of methane and 8.8 g of carbon dioxide in a 10 L vessel at 27° C is kPa. (Round off to the Nearest Integer). [Assume gases are ideal, R =8.314 J mol -1 K -1 Atomic masses: C: 12.0 u , H: 1.0 u , O: 16.0 u ] (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Total moles of gases, n = n CH 4+ n CO 2 =(6.4/16)+(8.8/44)=0.6 text Now, P =( nRT / V )=(0.6 × 8.314 × 300/10 × 10-3) =1.49652 × 105 Pa =149.652 kPa ≈ 150 kPa

Q. The pressure exerted by a non-reactive gaseous mixture of 6.4g of methane and 8.8g of carbon dioxide in a 10 L vessel at 27∘C is _______ kPa. (Round off to the Nearest Integer). [Assume gases are ideal, R=8.314Jmol−1K−1 Atomic masses : C:12.0u,H:1.0u,O:16.0u]

Total moles of gases, n=nCH4​​+nCO2​​ =166.4​+448.8​=0.6 Now, P=VnRT​=10×10−30.6×8.314×300​ =1.49652×105Pa=149.652kPa ≈150kPa

Total moles of gases, $n = n_{C H_{4}} + n_{C O_{2}}$

$= \frac{6.4}{16} + \frac{8.8}{44} = 0.6$

$Now, P = \frac{n RT}{V} = \frac{0.6 \times 8.314 \times 300}{10 \times 1 0 ^{- 3}}$

$= 1.49652 \times 1 0^{5} P a = 149.652 k P a$

$\approx 150 k P a$