The temperature of an open room of volume 30 m3 increases from 17°C to 27°C due to the sunshine. The atmospheric pressure in the room remains 1 × 105 Pa. If ni and nf are the number of molecules in the room before and after heating, then nf-ni will be :

Question

The temperature of an open room of volume 30 m3 increases from 17°C to 27°C due to the sunshine. The atmospheric pressure in the room remains 1 × 105 Pa. If ni and nf are the number of molecules in the room before and after heating, then nf-ni will be : (A)  - 1.61 × 1023 (B) 1.38 × 1023 (C) 2.5 × 1025 (D)  - 2.5 × 1025

Tardigrade Admin · Accepted Answer

n1 = initial number of moles n1 = (P1V1/RT1) = (105 ×30/8.3 ×290) ≈ 1.24 ×103 n2 = final number of moles =(P2V2/RT2) = (105 ×30/8.3 ×300) ≈ 1.20 ×103 Change of number of molecules: nf - ni = (n2 - n1)×6.023 ×1023 ≈ - 2.5 ×1025

$- 1.61 \times 1 0^{23}$

$1.38 \times 1 0^{23}$

$2.5 \times 1 0^{25}$

$- 2.5 \times 1 0^{25}$

Q. The temperature of an open room of volume 30m3 increases from 17∘C to 27∘C due to the sunshine. The atmospheric pressure in the room remains 1×105Pa. If ni​ and nf​ are the number of molecules in the room before and after heating, then nf​−ni​ will be :

−1.61×1023

1.38×1023

2.5×1025

−2.5×1025

n1​ = initial number of moles n1​=RT1​P1​V1​​=8.3×290105×30​ ≈1.24×103 n2​ = final number of moles =RT2​P2​V2​​=8.3×300105×30​ ≈1.20×103 Change of number of molecules : nf​−ni​=(n2​−n1​)×6.023×1023 ≈−2.5×1025

$- 1.61 \times 1 0^{23}$

$1.38 \times 1 0^{23}$

$2.5 \times 1 0^{25}$

$- 2.5 \times 1 0^{25}$