JEE WhatsApp Group
Sign in
Please select an account to continue using cracku.in
↓ →
JEE WhatsApp Group
Join Our JEE Preparation Group
Prep with like-minded aspirants; Get access to free daily tests and study material.
Half mole of an ideal monoatomic gas is heated at a constant pressure of 1 atm from 20$$^{\circ}$$C to 90$$^{\circ}$$C. Work done by the gas is
$$\Delta T = T_2 - T_1 = 90 - 20 = 70\text{ K}$$
$$W = P\Delta V$$
$$PV = nRT \implies P\Delta V = nR\Delta T$$
$$W = nR\Delta T$$
$$W = 0.5 \times \frac{25}{3} \times 70 = 35 \times \frac{25}{3} \approx 291\text{ J}$$
Create a FREE account and get:
Educational materials for JEE preparation
Ask our AI anything
AI can make mistakes. Please verify important information.
AI can make mistakes. Please verify important information.
