4.4 g of an ideal gas ( mol . wt =44) at 300 K and 10 atm is allowed is allowed to expand isothermally against a external pressure of 1 atm . Taking 1 L -atm =100 J , heat absorbed by the gas is:

Question

4.4 g of an ideal gas ( mol . wt =44) at 300 K and 10 atm is allowed is allowed to expand isothermally against a external pressure of 1 atm . Taking 1 L -atm =100 J , heat absorbed by the gas is: (A) 200.5 J (B) 212.5 J (C) 221.4 J (D) 242.4 J

Tardigrade Admin · Accepted Answer

Vi=(n R T/P)=(4.4/44) × (0.082 × 300/10) =0.246 litre V f =( P 1 V i / P f ) =(10 × 0.246/1)=2.46 litres q=-W=P(Vf-Vi) =1(2.46-0.246) × 100=221.4 J

Q. $4.4 g$ of an ideal gas $(m o l . wt = 44)$ at $300 K$ and $10 a t m$ is allowed is allowed to expand isothermally against a external pressure of $1 a t m .$ Taking $1 L$ -atm $= 100 J,$ heat absorbed by the gas is:

200.5 J

212.5 J

221.4 J

242.4 J

$V_{i} = \frac{n RT}{P} = \frac{4.4}{44} \times \frac{0.082 \times 300}{10}$

$= 0.246$ litre

$V_{f} = \frac{P _{1} V _{i}}{P _{f}}$

$= \frac{10 \times 0.246}{1} = 2.46$ litres

$q = - W = P (V_{f} - V_{i})$

$= 1 (2.46 - 0.246) \times 100 = 221.4 J$

Q. 4.4g of an ideal gas (mol.wt=44) at 300K and 10atm is allowed is allowed to expand isothermally against a external pressure of 1atm. Taking 1L -atm =100J, heat absorbed by the gas is: