A 5 , L cylinder contains 10 , mol of O_2 gas | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Initial volume $V_{initial} = 5 \, L$. Temperature $T = 27 + 273 = 300 \, K$. Final volume $V_{final}$ is the volume occupied by $10 \, mol$ of ga

Vedclass · Accepted Answer

$-1 	imes 241.3 \, L \cdot atm$

Vedclass · Answer

(C) Initial volume $V_{initial} = 5 \, L$. Temperature $T = 27 + 273 = 300 \, K$. Final volume $V_{final}$ is the volume occupied by $10 \, mol$ of gas at $1 \, atm$ pressure: $V_{final} = \frac{nRT}{P} = \frac{10 	imes 0.0821 	imes 300}{1} = 246.3 \, L$. Change in volume $\Delta V = V_{final} - V_{initial} = 246.3 - 5 = 241.3 \, L$. Work done $W = -P_{ext} \Delta V = -1 \, atm 	imes 241.3 \, L = -241.3 \, L \cdot atm$.

$A$ $5 \, L$ cylinder contains $10 \, mol$ of $O_2$ gas at $27 \, ^\circ C$. Due to a leak,the gas escapes into the atmosphere until the pressure inside the cylinder equals the atmospheric pressure $(1.0 \, atm)$. What is the work done by the gas?

Similar Questions

In thermodynamics,a process is called reversible when

An ideal gas is expanded from $1 \times 10^{-3} \,m^3$ to $1 \times 10^{-2} \,m^3$ against a constant external pressure of $1 \times 10^5 \,N/m^2$ at $300 \,K$. The work done is

Which law of thermodynamics provides the definition of temperature?

Explain pressure-volume work.

Which among the following pair of properties is intensive?

Vedclass Test Series

Exam Paper Generator

Online Exam Module